ESSENTIAL COHOMOLOGY AND EXTRASPECIAL ppp-GROUPS

Pham Anh Minh

Abstract. Let p be an odd prime number and let G be an extraspecial p-group. The purpose of the paper is to show that G has no non-zero essential mod-p cohomology (and in fact that H^{∗}(G, 𝔽_{p}) is Cohen-Macaulay) if and only if |G| = 27 and exp(G) = 3.

1. Introduction

Let p be a prime number. For every p-group K, denote by H^{∗}(K) the modp cohomology ring of K. A mod-p cohomology class of K is called essential if it vanishes on restriction to every proper subgroup of K. Let Ess(K) be the ideal of H^{∗}(K) consisting of such classes of H^{∗}(K). As observed in [3], the study of Ess(K) could provide interesting information for H^{∗}(K) (but, in contrast, it seems in general rather difficult to obtain elements of Ess(K)). For instance, Ess(K) ≠ {0} implies that the depth of H^{∗}(K) is just the rank of the center of K (see [3] and [5]); furthermore, with the condition that H^{∗}(K) is Cohen-Macaulay, it follows from [1] that Ess(K) ≠ {0} if and only if every element of order p of K is central (a way how to obtain some element of Ess(K) is this case was shown there).

We are now interested in extraspecial p-groups G. For p = 2, it was proved by Quillen ([17]) that H^{∗}(G) is Cohen-Macaulay and Ess(G) = {0}, except for the case G = Q_{8}, the quaternion group of order 8 (this fact also follows from Adem-Karagueuzian’s result, as Q_{8} is the unique group of which every element of order 2 is central). However, the situation is quite different for the case p > 2 - which is assumed from now on. Consider first the case |G| = p^{3}; it follows from [8], [9], [10],

[16] that Ess(G) ≠ {0} (so H^{∗}(G) is not Cohen-Macaulay) if and only if exp(G) > 3. In order to generalize this fact, in this note, we prove

Theorem. If G is an extraspecial p-group, then Ess(G) = {0} iff exp(G) = 3 and |G| = 3^{3}.

It follows that the unique extraspecial p-group which has no non-zero essential cohomology is the one of order 27 and of exponent 3. In each of the remaining cases, H^{∗}(G) is not Cohen-Macaulay and the depth of H^{∗}(G) is just 1; we also point out

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 20J06; Secondary 20D15, 55R40.

some non-zero essential classes of G (it turns out that, if |G| = p^{5} or exp(G) = p^{2}, there exists such a class of G belonging to Im Inf_{G}^{G/Z} with Z the center of G).

The note is organized as follows. In Section 2, given an extraspecial p-group G of order p^{2n+1}, we shall consider G as a subgroup of the central product Γ_{n} = C_{p^{2}} • G and give a sufficient and necessary condition for the fact that Res_{G}^{Γ_{n}}(ξ) ≠ 0 with ξ ∈ H^{∗}(Γ_{n}). The proofs of the theorem for the cases exp(G) > p or |G| = p^{5}, which are rather simple, will be given in Section 3. Section 4 is devoted to the case exp(G) = p.

2. The group Γ_{n}

Let us recall that an extraspecial p-group G is of order p^{2n+1} (n ∈ ℕ) and is isomorphic to one of the following central products of groups

\begin{aligned}𝔼_{n} & = 𝔼 • ⋯ • 𝔼 (n times), \\ 𝕄_{n} & = 𝕄 • 𝔼_{n−1}, \end{aligned}

where

\begin{aligned}𝕄 & = ⟨a, b|a^{p^{2}} = b^{p} = 1, b^{−1}ab = a^{1+p}⟩, \\ 𝔼 & = ⟨a, b|a^{p} = b^{p} = [a, b]^{p} = [a, [a, b]] = [b, [a, b]] = 1⟩ \end{aligned}

are extraspecial p-groups of order p^{3}. Note that

exp(G) = \{ \begin{matrix}p^{2}, & for G = 𝕄_{n} \\ p, & for G = 𝔼_{n}, \end{matrix}

and 𝕄_{n} = 𝕄_{n−1} • 𝕄.

These groups can be obtained cohomologically as follows. Let V be a vector space of dimension 2n+1 over the prime field 𝔽_{p} with basis e, a_{1}, …, a_{2n}. Let x, x_{1}, …, x_{2n} be a basis of H^{1}(V), dual to that of V and let y = βx, y_{i} = βx_{i} with β the Bockstein homomorphism, so

H = H^{∗}(V) = E[x, x_{1}, …, x_{2n}] ⊗ 𝔽_{p}[y, y_{1}, …, y_{2n}]

with E[u, v, …] (resp. 𝔽_{p}[u, v, …]) the exterior (resp. polynomial) algebra over 𝔽_{p} with generators u, v, … of degree 1 (resp. 2). Consider the central extension

(Γ_{n})  0 → 𝔽_{p} →^{i} Γ_{n} → V → 0,

with factor set z = z_{n} = y + x_{1}x_{2} + ⋯ + x_{2n−1}x_{2n}. Via the inflation map, x and the x_{i}’s can be considered as elements of H^{1}(Γ_{n}). Given a subgroup K of Γ_{n}, with some abuse of notation, we also denote by x (resp. x_{i}) the element Res_{K}^{Γ_{n}}(x) (resp. Res_{K}^{Γ_{n}}(x_{i})).

ESSENTIAL COHOMOLOGY AND EXTRASPECIAL p p p-GROUPS 3

It is easy to show Lemma 1. (i) Γ_{n} = C_{p^{2}} • 𝕄_{n} = C_{p^{2}} • 𝔼_{n} = Γ_{n−1} • 𝕄;

(ii)  𝕄_{n} = Ker (x + α),
𝔼_{n} = Ker x,
Γ_{n−1} × C_{p} = Ker α,

with α a non-zero linear combination of x_{1}, …, x_{2n}. □

C_{p^{2}} = ∩^{2n}_{i=1}Ker x_{i} is then a subgroup of Γ_{n}. Let w be a generator of H^{2}(C_{p^{2}}), so

H^{∗}(C_{p^{2}}) = E[x] ⊗ 𝔽_{p}[w].

Set G_{n} = C_{p^{2}} × 𝔼_{n}. By Künneth formula, we have

H^{∗}(G_{n}) = H^{∗}(𝔼_{n}) ⊗ E[x] ⊗ 𝔽_{p}[w].

As Γ_{n} is the central product of C_{p^{2}} and 𝔼_{n}, there exists a central subgroup U_{n} of order p of G_{n} such that G_{n}/U_{n} = Γ_{n} and the factor set of the central extension

1 → U_{n} → G_{n} → Γ_{n} → 1

is just y. Consider the following commutative diagram

(1)  \begin{aligned}1 & ⟶ & U_{n} & ⟶ & U_{n} × 𝔼_{n} & ⟶ & 𝔼_{n} & ⟶ & 1 \\ & & ∥ & & ↓ & & ↓ \\ 1 & ⟶ & U_{n} & ⟶ & G_{n} & ⟶ & Γ_{n} & ⟶ & 1 \end{aligned}  ,

where rows are central extensions and vertical arrows are inclusion maps. Pick elements s, t of H^{∗}(U_{n}) satisfying H^{∗}(U_{n}) = E[s] ⊗ P [t]. It follows from [11] (see also [2]) that t can be chosen so that Res_{U_{n}}^{G_{n}}(w × 1) = t.

We now use the following notation. Given a ring R and elements r, s, ⋯ ∈ R, (r, s, …) will denote the ideal of R generated by r, s, . . .. The main result of this section is the following

Proposition 1. Given ξ ∈ H^{∗}(Γ_{n}), then Res_{𝔼_{n}}^{Γ_{n}}(ξ) ≠ 0 iff xξ ∉ (y).

Proof. Set X =Inf_{G_{n}}^{Γ_{n}}(ξ). As Ker Inf_{G_{n}}^{Γ_{n}} = (y), it follows that xX = 0 iff xξ ∈ (y). Write X = ∑ w^{i} ⊗ s_{i} + ∑ w^{i}x ⊗ t_{i} with s_{i}, t_{i} ∈ H^{∗}(𝔼_{n}). It is clear that Res_{𝔼_{n}}^{Γ_{n}}(ξ) ≠ 0 iff Inf_{U_{n}×𝔼_{n}}^{𝔼_{n}}Res_{𝔼_{n}}^{Γ_{n}}(ξ) ≠ 0. So, by the commutative diagram (1), Res_{𝔼_{n}}^{Γ_{n}}(ξ) ≠ 0 iff Res_{U_{n}×𝔼_{n}}^{G_{n}}(X) ≠ 0, which is equivalent to the fact the s_{i}’s are not all equal to zero, or equivalently, xξ ∉ (y). The proposition follows. □

For convenience, given a central extension of groups

(K)  1 → A → K → C → 1,

denote by {E_{r}(K), d_{r}} the Hochschild-Serre spectral sequence corresponding to the extension (K). We now recall some results given in [12], [13] (see also [2] for n = 1) concerning {E_{r}(Γ_{n}), d_{r}}. As usual, denote by P^{i} the Steenrod operations. Set Z = i(𝔽_{p}) ⊂ C_{p^{2}} ⊂ Γ_{n}. So v = Res_{Z}^{C_{p^{2}}} (w) is a generator of H^{2}(Z). Let

X_{n} = x_{1}x_{2} … x_{2n−1}x_{2n},
η_{i} = P^{p^{i−2}} … P^{1}βz

= ∑_{j=1}^{n}(x_{2j−1}y_{2j}^{p^{i−1}} − x_{2j}y_{2j−1}^{p^{i−1}}),

ξ_{m} = βP^{p^{m−1}} … P^{1}βz

= ∑_{j=1}^{n}(y_{2j−1}y_{2j}^{p^{m}} − y_{2j}y_{2j−1}^{p^{m}}),

1 ≤ i ≤ n + 1, 1 ≤ m ≤ n, be elements of H^{∗}(V). We have

Theorem 1 ([2], as corrected in [13, Rk. 2.11(ii)], [12]).

(i)

\begin{matrix}E_{∞}(Γ_{1}) = & H^{∗}(V)/(z, η_{1}, η_{2}, ξ_{1}) ⊗ 𝔽_{p}[v^{p}] \\ ⊕ (𝔽_{p}X_{1} ⊕ 𝔽_{p}xX_{1}) ⊗ ∑_{i=1}^{p−2} 𝔽_{p}[v^{p}]v^{i}. \end{matrix}

(ii) For n ≥ 2,

\begin{matrix}E_{2p+1}(Γ_{n}) = & H^{∗}(V)/(z, η_{1}, ξ_{1}, A_{n}η_{2}) ⊗ 𝔽_{p}[v^{p}] \\ ⊕ (𝔽_{p}X_{n} ⊕ 𝔽_{p}xX_{n}) ⊗ ∑_{i=1}^{p−2} 𝔽_{p}[v^{p}]v^{i} \end{matrix}

with A_{n} = ∑^{n} ; i=1x_{1}x_{2} … x̂_{2i−1}x̂_{2i} … x_{2n−1}x_{2n}. □

Let W be the vector subspace of V given by W = Ker (x − x_{1}). We have then the central extension

(𝕄_{n})  1 → Z → 𝕄_{n} → W → 0

with factor set z′ = z′_{n} = y_{1} + x_{1}x_{2} + ⋯ + x_{2n−1}x_{2n}. Following [10], [12], we also have Proposition 2 ([10], [12]).

(i)

E_{∞}(𝕄_{1}) = E_{2p+1}(𝕄_{1}) =H^{∗}(W)/(z′, βz′) ⊗ 𝔽_{p}[v^{p}]

⊕ (𝔽_{p}[y_{2}]x_{1} ⊕ 𝔽_{p}[y_{2}]x_{1}x_{2}) ⊗ 𝔽_{p}[v^{p}]v^{p−1}

⊕ (𝔽_{p}x_{1} ⊕ 𝔽_{p}x_{1}x_{2}) ⊗ ∑_{i=1}^{p−2} 𝔽_{p}[v^{p}]v^{i}.

(ii) For n ≥ 2,

\begin{matrix}E_{2p+1}(𝕄_{n}) = & H^{∗}(W)/(z′, βz′, βP^{1}βz′, B_{n} · P^{1}βz′) ⊗ 𝔽_{p}[v^{p}] \\ ⊕ (𝔽_{p}x_{1}x_{3}x_{4} … x_{2n−1}x_{2n} ⊕ 𝔽_{p}X_{n}) ⊗ ∑_{i=1}^{p−2} 𝔽_{p}[v^{p}]v^{i} \end{matrix}

with B_{n} = ∑^{n} ; i=2x_{1}x_{3}x_{4} … x̂_{2i−1}x̂_{2i} … x_{2n−1}x_{2n}. □

We also prove

Proposition 3. If ξ ∈ H^{∗}(Γ_{n}) and |ξ| < 2n + 2, then ξ ∈ Im Inf_{Γ_{n}}^{V}.

The proof of the proposition is divided into the following lemmas. Set

R = E[x_{1}, …, x_{m}] ⊗ 𝔽_{p}[t_{1}, …, t_{m}],
and  α_{i} = ∑_{j=1}^{m} x_{j}t_{j}^{p^{i−1}}, 1 ≤ i ≤ m,

be elements of R. Denote by I_{k,m} the set consisting of subsets of k elements of {1, …, m}. For every element I = {i_{1}, …, i_{k}} of I_{k,m} with i_{1} < ⋯ < i_{k}, set x_{I} = x_{i_{1}} … x_{i_{k}}and x_{∅} = 1.

Lemma 2. For X ∈ R and 1 ≤ k ≤ m, (i) if X · α_{1} … α_{k} = 0, then X ∈ (α_{1}, …, α_{k}, x_{I}|I ∈ I_{m−k+1,m}); (ii) if X · α_{k} = 0, then X ∈ (α_{k}, x_{1} … x_{m}).

Proof. (i) By induction on m. The case m = 2 is obvious. Assume that (i) holds for m − 1.

If k = m, then

α_{1} … α_{m} = | \begin{matrix}t_{1} & … & t_{m} \\ t_{1}^{p} & … & t_{m}^{p} \\ · & … & · \\ t_{1}^{p^{m−1}} & … & t_{m}^{p^{m−1}} \end{matrix} | x_{1} … x_{m};

so X ∈ (x_{i}|1 ≤ i ≤ m). Suppose that k < m. Write

α_{i} = α′_{i} + x_{m}t_{m}^{p^{i−1}}

with α′_{i} = ∑_{j=1}^{m−1} x_{j}t_{j}^{p^{i−1}}, 1 ≤ i ≤ m, and

X = X′ + X′′x_{m},

with X′, X′′ free of x_{m}. Since Xα_{1} … α_{k} = 0, we have

0 = X′α′_{1} … α′_{k},
and  0 = (−1)^{k}X′′α′_{1} … α′_{k} + X′ ∑_{i=1}^{k}(−1)^{k−i}t_{m}^{p^{i−1}}α′_{1} … α′_{i} … α′_{k}.

By writing

\begin{aligned}X′ & = t_{m}^{r_{1}}f_{1} + ⋯ + t_{m}^{r_{j}}f_{j}, \\ X′′ & = t_{m}^{s_{1}}g_{1} + ⋯ + t_{m}^{s_{i}}g_{i} \end{aligned}

with f_{i}, g_{j} free of t_{m}, r_{1} < ⋯ < r_{j}, s_{1} < ⋯ < s_{i}, we have

(2)  (−1)^{k}t_{m}^{s_{i}}g_{i}α′_{1} … α′_{k} + t_{m}^{p^{k−1}+r_{j}}f_{j}α′_{1} … α′_{k−1} = 0.

Consider the following cases:

- r_{j} + p^{k−1} > s_{i}: from (2), f_{j}α′_{1} … α′_{k−1} = 0. By the inductive hypothesis, f_{j} ∈ (α′_{1}, …, α′_{k−1}, I_{m−k+1,m−1}). Since α′_{i} = α_{i} −x_{m}t_{m}^{p^{i−1}}, we have X = t_{m}^{r_{1}}f_{1} +⋯+ t_{m}^{r_{j−1}}f_{j−1} mod (x_{m}, α_{1}, …, α_{k}, I_{m−k+1,m}). So we may suppose that f_{j} = 0.

- r_{j} + p^{k−1} < s_{i}: from (2), g_{i}α′_{1} … α′_{k} = 0. By the inductive hypothesis, g_{i} ∈ (α′_{1}, …, α′_{k}, I_{m−k,m−1}). So x_{m}g_{i} ∈ (α_{1}, …, α_{k}, I_{m−k+1,m}).

- r_{j} + p^{k−1} = s_{i}: from (2), ((−1)^{k+k−1}g_{i}α′_{k} + f_{j})α′_{1} … α′_{k−1} = 0. By the inductive hypothesis, f_{j} = g_{i}α′_{k} mod (α′_{1}, …, α′_{k−1}, I_{m−k+1,m−1}). Since α′_{i} = α_{i} − x_{m}t_{m}^{p^{i−1}}, we have f_{j} = g_{i}α′_{k} mod (x_{m}, α_{1}, …, α_{k−1}, I_{m−k+1,m}). So we may suppose that f_{j} = g_{i}α′_{k}. Since

t_{m}^{r_{j}}f_{j} + t_{m}^{s_{i}}g_{i}x_{m} = t_{m}^{r_{j}}(f_{j} + t_{m}^{p^{k−1}}g_{i}x_{m})
= t_{m}^{r_{j}}g_{i}(α′_{k} + t_{m}^{p^{k−1}}x_{m})
= t_{m}^{r_{j}}g_{i}α_{k}.

We may then suppose that f_{j} = 0 and g_{i} = 0.

The above arguments show that we may reduce to the case X′ = 0. It follows that X′′α′_{1} … α′_{k} = 0. By the inductive hypothesis, X′′ ∈ (α′_{1}, …, α′_{k}, I_{m−k,m−1}). Hence x_{m}X′′ ∈ (α_{1}, …, α_{k}, I_{m−k+1,m}). (i) is proved.

(ii) By induction on m. The case m = 1 is trivial. Assume that (ii) holds for m − 1. As above, write

X = X′ + X′′x_{m},

with X′, X′′ free of x_{m}. Arguing as above, we may reduce to the case X′ = 0. It follows that X′′α′_{k} = 0. Bu the inductive hypothesis, X′′ ∈ (α′_{k}, x_{1} … x_{m−1}). Hence x_{m}X′′ ∈ (α_{k}, x_{1} … x_{m}).

The lemma is proved. □

Lemma 3. Let 1 ≤ k ≤ n and let Y_{1}, …, Y_{k} be elements of H^{∗}(V).

(i) If Y_{1}ξ_{1} + ⋯ + Y_{k}ξ_{k} = 0, then Y_{k} ∈ (ξ_{1}, …, ξ_{k−1}); (ii) Assume that

Y_{k} = _{∑_{I⊂{1,...,2n}} ; #(I)<2n−k+1}x_{I}f_{I}(y, y_{1}, …, y_{2n}).

We have

(iia) if Y_{1}η_{1} + ⋯ + Y_{k}η_{k} = 0, then Y_{k} ∈ (η_{1}, …, η_{k});

(iib) if Y_{k} ∈ ∩^{k}_{i=1}(η_{i}), then Y_{k} ∈ (η_{1} … η_{k}).

(iic) if Y_{1}ξ_{1} + ⋯ + Y_{k−1}ξ_{k−1} + Y_{k}η_{ℓ} = 0 with 1 ≤ ℓ ≤ n, then Y_{k} ∈

(η_{ℓ}, ξ_{1}, …, ξ_{k−1}).

Proof. (i) For 1 ≤ i ≤ k, write

Y_{i} = ∑_{I⊂{1,...,2n}} x_{I}f_{I}^{(i)}(y, y_{1}, …, y_{2n}).

Then, for every I, we have

∑_{i=1}^{k} f_{I}^{(i)}ξ_{i} = 0.

According to [18], ξ_{1}, …, ξ_{k} is a regular sequence in P. So the above equality implies f_{I}^{(k)} ∈ (ξ_{1}, …, ξ_{k−1}). Therefore Y_{k} ∈ (ξ_{1}, …, ξ_{k−1}).

(iia) It follows that Y_{k}η_{1} … η_{k} = 0. By Lemma 2, Y_{k} ∈ (η_{1}, …, η_{k}, I_{2n−k+1,2n}). So Y_{k} ∈ (η_{1}, …, η_{k}).

(iib) By induction on k. For k = 2, Xη_{1} + Y η_{2} = 0 implies Y η_{1}η_{2} = 0. By Lemma 2, Y ∈ (η_{1}, η_{2}, I_{2n−1,2n}). So Y ∈ (η_{1}, η_{2}). Write Y = aη_{1} + bη_{2}. Then Y_{2} = Y η_{2} = aη_{1}η_{2}.

Assume that (iib) holds for k − 1 ≥ 2. As Y_{k} ∈ ∩_{i<k}(η_{i}), it follows from the inductive hypothesis that Y_{k} = Y η_{1} … η_{k−1}. Write Y_{k} = Xη_{k}. Then Y η_{1} … η_{k} = 0. By Lemma 2, Y = c_{1}η_{1} + ⋯ + c_{k}η_{k}. So Y_{k} = (−1)^{k−1}c_{k}η_{1} … η_{k}.

(iic) By induction on k. Y_{1}ξ_{1} +Y_{2}η_{ℓ} = 0 implies Y_{1}η_{ℓ} = 0. By Lemma 2, Y_{1} ∈ (η_{ℓ}). Write Y_{1} = cη_{ℓ}. Then (cξ_{1} + Y_{2})η_{ℓ} = 0. By Lemma 2, cξ_{1} + Y_{2} ∈ (η_{ℓ}), hence Y_{2} ∈ (η_{ℓ}, ξ_{1}).

Assume that (iic) holds for k − 1 ≥ 2. As Y_{1}η_{ℓ}ξ_{1} + ⋯ + Y_{k−1}η_{ℓ}ξ_{k−1} = 0, it follows from (i) that Y_{k−1}η_{ℓ} ∈ (ξ_{1}, …, ξ_{k−2}). By the inductive hypotheses, we may write Y_{k−1} = c_{1}ξ_{1} + ⋯ + c_{k−2}ξ_{k−2} + c_{k−1}η_{ℓ}. Hence

(Y_{1} + c_{1}ξ_{k−1})ξ_{1} + ⋯ + (Y_{k−2} + c_{k−2}ξ_{k−1})ξ_{k−2} + (Y_{k} + c_{k−1}ξ_{k−1})η_{ℓ} = 0.

By the inductive hypothesis, Y_{k} + c_{k−1}ξ_{k−1} ∈ (ξ_{1}, …, ξ_{k−2}, η_{ℓ}), hence Y_{k} ∈ (ξ_{1}, …, ξ_{k−1}, η_{ℓ}). □

For 1 ≤ i ≤ n + 1, 0 ≤ k ≤ n, denote by Δ_{i,k} the ideal of H^{∗}(V) given by

Δ_{i,k} = \{ \begin{matrix}(z, η_{j}, ξ_{m}|1 ≤ j ≤ i, 1 ≤ m ≤ k) & if k ≥ 1, \\ (z, η_{j}|1 ≤ j ≤ i) & if k = 0. \end{matrix}

Lemma 4. If X = ∑_{#(I)<2n−2k+1}x_{I}X_{I}(y_{1}, …, y_{2n}) and Xξ_{j} ∈ Δ_{k,j−1} with 1 ≤ j ≤

k ≤ n, then X ∈ Δ_{k,j−1}.

Proof. Write

Xξ_{j} = a_{0}z + ∑_{i=1}^{k} a_{i}η_{i} + ∑_{i=1}^{j−1} b_{i}ξ_{i}

with a_{i}, b_{ℓ} ∈ H^{∗}(V). Since y = z − x_{1}x_{2} − ⋯ − x_{2n−1}x_{2n}, we may suppose that a_{i}, b_{ℓ}, 1 ≤ i ≤ k, 1 ≤ ℓ ≤ j − 1, are free of y. It follows that a_{0} = 0 and

(3)  Xξ_{j}η_{1} … η_{k} = ∑_{i=1}^{j−1} b_{i}ξ_{i}η_{1} … η_{k}.

We now argue by induction on j. For j = 1, it follows that Xξ_{1}η_{1} … η_{k} = 0. Hence Xη_{1} … η_{k} = 0. By Lemma 2, X ∈ (η_{1}, …, η_{k}).

Assume that the lemma holds for j − 1 ≥ 1. By Lemma 3 (i) and by (3), there exists c_{i} ∈ H^{∗}(V) such that

Xη_{1} … η_{k} = c_{1}ξ_{1} + ⋯ + c_{j−1}ξ_{j−1}.

Therefore, by Lemma 3 (i), c_{j−1}η_{i} ∈ (ξ_{1}, …, ξ_{j−2}), for every 1 ≤ i ≤ k; by Lemma 3 (iic), c_{j−1} ∈ ∩_{i≤k}(ξ_{1}, …, ξ_{j−2}, η_{i}). By writing

\begin{aligned}c_{j−1} & = d_{1}ξ_{1} + ⋯ + d_{j−2}ξ_{j−2} + dη_{1} … η_{i−1} \\ & = e_{1}ξ_{1} + ⋯ + e_{j−2}ξ_{j−2} + eη_{i}, \end{aligned}

we get

[(e_{1} − d_{1})ξ_{1} + ⋯ + (e_{j−2} − d_{j−2})ξ_{j−2}]η_{1} … η_{i} = 0.

By Lemma 2, (e_{1} −d_{1})ξ_{1} +⋯+(e_{j−2} −d_{j−2})ξ_{j−2} contains η_{1} … η_{i} as a factor. Hence eη_{i} ∈ ∩_{ℓ≤i}(η_{ℓ}). By Lemma 3 (iib), c_{j−1} ∈ (ξ_{1}, …, ξ_{j−2}, η_{1} … η_{i}). So we may suppose that c_{j−1} ∈ (η_{1} … η_{k}). By writing c_{j−1} = cη_{1} … η_{k}, we have

(X − cξ_{j−1})η_{1} … η_{k} = c_{1}ξ_{1} + ⋯ + c_{j−2}ξ_{j−2}.

By the inductive hypothesis, this implies X − cξ_{j−1} ∈ Δ_{k,j−2}. So X ∈ Δ_{k,j−1}. The lemma follows. □

Lemma 5. If X = ∑_{#(I)<2n−2k}x_{I}X_{I}(y_{1}, …, y_{2n}) and Xη_{k} ∈ Δ_{k−1,k−1} with 1 ≤ k ≤

n + 1, then X ∈ Δ_{k,k−1}.

Proof. Write

Xη_{k} = a_{0}z + ∑_{i=1}^{k−1}(a_{i}η_{i} + b_{i}ξ_{i}).

Arguing as in the proof of Lemma 4, we may suppose that a_{i}, b_{i}, with 1 ≤ i ≤ k − 1, are free of y. It follows that a_{0} = 0 and

Xη_{k} = ∑_{i=1}^{k−1}(a_{i}η_{i} + b_{i}ξ_{i}).

Furthermore, we may suppose that every b_{i} is of form

b_{i} = ∑_{#(I)<2n−2k+1} x_{I}b_{I}^{(i)}.

Therefore, applying Lemma 4 yields b_{k−1} ∈ Δ_{k,k−2}. Hence, by induction, we need only consider the case

Xη_{k} = b_{1}ξ_{1} + ∑_{i=1}^{k−1} a_{i}η_{i}.

This implies b_{1}ξ_{1}η_{1} … η_{k} = 0. So b_{1}η_{1} … η_{k} = 0. By Lemma 2, b_{1} ∈ (η_{1}, …, η_{k}). The lemma follows. □

Let us now consider the Hochschild-Serre spectral sequence {E_{r}(Γ_{n}), d_{r}}. It follows that, for k < 2n + 2,

∑_{i+j=k} E_{2p+1}^{i,j} ⊂ E_{2p+1}^{k,0} ⊕ ⊕_{r≥1}E_{2p+1}^{∗,2pr}.

By Kudo’s transgression theorem, for m ≤ n, 1 ⊗ v^{p^{m}} (resp. η_{m} ⊗ v^{p^{m−1}(p−1)}) survives to E_{2p^{m}+1} (resp. E_{2p^{m−1}(p−1)+1}) and

\begin{aligned}d_{2p^{m}+1}(1 ⊗ v^{p^{m}}) & = η_{m+1}, \\ d_{2p^{m−1}(p−1)+1}(η_{m} ⊗ v^{p^{m−1}(p−1)}) & = −ξ_{m}. \end{aligned}

Lemma 6. For k < 2n + 2 and 1 ≤ m ≤ n, we have

∑_{i+j=k} E_{2p^{m}+1}^{i,j} ⊂ E_{2p^{m}+1}^{k,0} ⊕ ⊕_{r≥1}E_{2p^{m}+1}^{∗,2p^{m}r}.

Proof. By the structure of E_{2p+1}(Γ_{n}), the lemma holds for m = 1. Suppose that the lemma holds for m = s ≥ 1. Let ψ = X ⊗ v^{ℓp^{s}} be an element of E_{2}(Γ_{n}) surviving to E_{2p^{s+1}+1}, with 1 ≤ ℓ ≤ p − 1 and |X| + 2ℓp^{s} = k < 2n + 2. So d_{2p^{s}+1}(ψ) = ℓXη_{s+1} ⊗ v^{(ℓ−1)p^{s}} must be hit by images under the differentials of elements of degrees less than 2n + 2. By the inductive hypothesis and by Kudo’s theorem, these images belong to the ideal Δ_{s,s}, hence so does Xη_{s+1}. Since, in E_{3}(Γ_{n}), y = −(x_{1}x_{2} + ⋯ + x_{2n−1}x_{2n}), we may suppose that X is free of y. As |X| < 2n + 2 − 2p^{s} < 2n − 2s − 2, by Lemma 5, this means that X ∈ Δ_{s+1,s}. So ψ = 0 in E_{2p^{s}+2} if ℓ < p − 1. If ℓ = p − 1, write ψ = Y η_{s+1} ⊗ v^{p^{s}(p−1)}. Then d_{2p^{s}(p−1)+1}(ψ) = −Y ξ_{s+1} ∈ Δ_{s+1,s}. Arguing as above, we may suppose that Y is free of y. By Lemma 4, as |ψ| < 2n + 2, Y ∈ Δ_{s+1,s}. So ψ = 0 in E_{2p^{s}(p−1)+2}. The lemma follows. □

Proof of Proposition 3. It follows from Lemma 6 that ξ either belongs to Im Inf_{Γ_{n}}^{V} or represents an element of E_{∞}^{∗,2p^{n}r}. As 2p^{n} > 2n + 2, the fact that |ξ| < 2n + 2 implies ξ ∈ Im Inf_{Γ_{n}}^{V}. The proposition follows. □

3. The case exp(G) > p or |G| = p^{5}

We first consider the case G = 𝕄_{n}. Consider G as a subgroup of Γ_{n} by setting G = Ker (x − x_{1}). If n = 1, it follows from [10] (see also [4]) that H^{∗}(𝕄) contains a nonzero essential element, namely X_{1}. Assume inductively that 0 ≠ X_{n−1} ∈ Ess(𝕄_{n−1}). As 𝕄_{n} = 𝕄_{n−1} • 𝕄, we have the following central extension

0 → 𝔽_{p} → 𝕄_{n−1} × 𝕄 → 𝕄_{n} → 1.

The fact that H^{∗}(𝕄_{n}) contains non-zero essential elements follows from

Proposition 4. 0 ≠ X_{n} ∈ Ess(𝕄_{n}).

Proof. Let K be a maximal subgroup of 𝕄_{n}. As dim_{𝔽_{p}}H^{1}(K) = 2n − 1, it follows that the product of any 2n elements of H^{1}(K) vanishes. Hence Res_{K}^{𝕄_{n}}(X_{n}) = 0 which implies that X_{n} ∈ Ess(𝕄_{n}). Furthermore, as Inf_{𝕄_{n−1}×𝕄}^{𝕄_{n}}(X_{n}) = X_{n−1} × x_{2n−1}x_{2n} ≠ 0 in H^{∗}(𝕄_{n−1} × 𝕄) by the inductive hypothesis, it follows that X_{n} ≠ 0. The proposition is proved □

By Theorem 1 (i), X_{1} and xX_{1} are non-zero elements of H^{∗}(Γ_{1}). By considering the central extension 0 → 𝔽_{p} → Γ_{n−1} × 𝕄 → Γ_{n} → 1, and by using the same argument given in the proof of Proposition 4, we also have

Proposition 5. The elements xX_{n} and X_{n} are non-zero elements of H^{∗}(Γ_{n}). □

Our next task is to prove that the theorem holds for the extraspecial p-group G = 𝔼_{2}. Consider 𝔼_{2} as a subgroup of Γ_{2} as in Lemma 1. Let Q be the element of H^{∗}(V) defined by Q = Q_{2,1}^{1,2} − Q_{2,1}^{3,4} with

Q_{2,1}^{i,j} = Q_{2,1}(y_{i}, y_{j}) = \frac{y_{i}^{p^{2}}y_{j} − y_{j}^{p^{2}}y_{i}}{y_{i}^{p}y_{j} − y_{j}^{p}y_{i}}

(so Q_{2,1}^{i,j} is nothing but the Dickson invariant of order 2(p^{2} − p) with variables y_{i}, y_{j}), and set η = x_{1}x_{2}Q. It follows from [19, Th. 8.25] that 0 ≠ η ∈ H^{∗}(𝔼_{2}). The case G = 𝔼_{2} is then proved by the following

Proposition 6. η ∈ Ess(𝔼_{2}).

Proof. Let K be a maximal subgroup of 𝔼_{2}, so K ≅ 𝔼 × C_{p}. If Res_{K}^{𝔼_{2}}(x_{3}x_{4}) = 0, it is clear that Res_{K}^{𝔼_{2}}(x_{1}x_{2}Q_{2,1}^{1,2}) = 0; we can then assume that Res_{K}^{𝔼_{2}}(x_{3}x_{4}) ≠ 0. Choose a basis u_{1}, u_{2}, u_{3}, u_{4} of H^{1}(𝔼_{2}/Z) such that K =Ker u_{4}, x_{1}x_{2} + x_{3}x_{4} = u_{1}u_{2} + u_{3}u_{4}, Res_{K}^{𝔼_{2}}(x_{1}x_{2}) = u_{1}u_{2} + u_{1}u_{3} and Res_{K}^{𝔼_{2}}(x_{3}x_{4}) = −u_{1}u_{3}. This implies that u_{1}u_{2} = 0 in H^{∗}(K). By setting v_{i} = βu_{i}, we have

\begin{aligned}Res_{K}^{𝔼_{2}}(η) & = (u_{1}u_{2} + u_{1}u_{3})(Q_{2,1}(v_{1}, v_{2} + v_{3}) − Q_{2,1}(v_{3}, v_{1})) \\ & = u_{1}u_{3}(Q_{2,1}(v_{1}, v_{2} + v_{3}) − Q_{2,1}(v_{3}, v_{1})) as u_{1}u_{2} = 0 in H^{∗}(K). \end{aligned}

Set Y = Q_{2,1}(v_{1}, v_{2} + v_{3}) − Q_{2,1}(v_{3}, v_{1}). Following [13, Proof of Lemma 1.10], Y contains v_{2}^{p} − v_{2}v_{1}^{p−1} as a factor. As u_{1}(v_{2}^{p} − v_{2}v_{1}^{p−1}) = −P^{1}β(u_{1}u_{2}) + v_{1}^{p−1}β(u_{1}u_{2}), we have Res_{K}^{𝔼_{2}}(η) = 0. So η ∈ Ess(𝔼_{2}). The proposition is proved. □

For exp(G) > p or |G| = p^{5}, Propositions 4 and 6 tell us that there exist non-zero essential cohomology classes of G which belong to Im Inf_{G}^{V}. Furthermore, if G = 𝕄_{2} then [12, Proposition 1.9] and [13, Theorem 3.10] tell us that x_{3}x_{4}N, (y_{3}x_{4} − y_{4}x_{3})N are also non-zero elements of Ess(𝕄_{2}) with N = (y_{2}^{p−1} − y_{3}^{p−1})(y_{2}^{p−1} − y_{4}^{p−1}). We can then end the section by the following

Question. For G ≅ 𝔼̸, is it true that Ess(G)∩ Im Inf_{G}^{V} ≠ {0} ?

4. The case exp(G) = p

We first point out some mod-p cohomology classes of Γ_{n}, by using the following argument given by D.J. Green [6]. Let K be a p-group containing C as a central subgroup. We have the central extension

(K)  1 → C → K →^{pr} K/C → 1,

On the other hand, by considering the extension

(K × C)  1 → C →^{ℓ} K × C →^{j} K → 1

with ℓ(c) = (1, c), j(k, c) = k, c ∈ C, k ∈ K, we have the commutative diagram

(4)  \begin{aligned}1 & ⟶ & C & ⟶^{ℓ} & K × C & ⟶^{j} & K & ⟶ & 1 \\ & & ∥ & & ^{μ}↓ & & ↓^{pr} \\ 1 & ⟶ & C & ⟶ & K & ⟶^{pr} & K/C & ⟶ & 1 \end{aligned}

with μ(k, c) = kc, k ∈ K, c ∈ C. The Hochschild-Serre spectral sequences corresponding to these extensions are of forms

\begin{aligned}E_{2}(K) & = H^{∗}(K/C) ⊗ H^{∗}(C) ⇒ H^{∗}(K), \\ E_{2}(K × C) & = E_{∞}(K × C) = H^{∗}(K) ⊗ H^{∗}(C). \end{aligned}

Furthermore, vertical arrows in (4) also induce a map {μ_{r} : E_{r}(K) → E_{r}(K × C)} between spectral sequences with μ_{2} = (Inf_{K}^{K/C}, 1_{H^{∗}(C)}).

The following is due to D.J. Green.

Proposition 7. For r ≥ 2, Im (d_{r} : E_{r}(K) → E_{r}(K)) ⊂ Ker Inf_{K}^{K/C} ⊗ H^{∗}(C).

Proof. Let ξ ∈ E_{r}(K) and write d_{r}(ξ) = ∑ ϕ_{j} ⊗ψ_{j}, ϕ_{j} ∈ H^{∗}(K/C), ψ_{j} ∈ H^{∗}(C). We can suppose that the ψ_{j}’s are linearly independent in H^{∗}(C). From the commutative diagram (4) and from the fact that d_{r} : E_{r}(K × C) → E_{r}(K × C) vanishes, we have

∑ Inf_{K}^{K/C}(ϕ_{j}) ⊗ ψ_{j} = μ_{r}(d_{r}(ξ)) = d_{r}(μ_{r}(ξ)) = 0.

So ϕ_{j} ∈ Ker Inf_{K}^{K/C}. The proposition follows. □

Since d_{2p+1}(v^{p}) = η_{2} in E_{2p+1}(Γ_{n}) (resp. P^{1}βz′ in E_{2p+1}(𝕄_{n})), it follows from Theorem 1 and Proposition 2 that A_{n} ⊗ v^{p} ∈ E_{2p+2}(Γ_{n}) and B_{n} ⊗ v^{p} ∈ E_{2p+2}(𝕄_{n}). We get then

Proposition 8. For 1 ≤ i ≤ p − 2,

(i) if n ≥ 2 then x_{1}x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{i}, X_{n} ⊗ v^{i}, x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{p} and B_{n} ⊗ v^{p} represent non-zero elements of E_{∞}(𝕄_{n});

(ii) X_{n} ⊗ v^{i}, xX_{n} ⊗ v^{i} and A_{n} ⊗ v^{p} represent non-zero elements of E_{∞}(Γ_{n}).

Proof. Note that, in H^{∗}(W), we have

x_{3}x_{4} … x_{2n−1}x_{2n} · P^{1}βz′ = (x_{3}x_{4} … x_{2n−1}x_{2n})(y_{1}^{p}x_{2} − y_{2}^{p}x_{1})
= (x_{3}x_{4} … x_{2n−1}x_{2n})(z′^{p}x_{2} + y_{2}^{p−1}βz′ − y_{2}^{p−1}y_{1}x_{2})
= (x_{3}x_{4} … x_{2n−1}x_{2n})(z′^{p}x_{2} + y_{2}^{p−1}βz′ − y_{2}^{p−1}z′x_{2})
∈ (z′, βz′).

So d_{2p+1}(x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{p}) = 0. Therefore x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{p} survives to E_{∞}(𝕄_{n}).

By Proposition 4, X_{n} ≠ 0 in H^{∗}(𝕄_{n}) implies that X_{n}, x_{1}x_{3}x_{4} … x_{2n−1}x_{2n} and B_{n} are not elements of Ker Inf_{𝕄_{n}}^{W}. Similarly, Proposition 5 shows that X_{n}, xX_{n} and A_{n} are not elements of Ker Inf_{Γ_{n}}^{V}. The proposition follows from Proposition 7. □

For n ≥ 1 and for 1 ≤ i ≤ p − 2, let us pick elements X_{n,i} ∈ H^{2(n+i)−1}(𝕄_{n}) and Y_{n,i} ∈ H^{2(n+i)}(Γ_{n}) which represent respectively x_{1}x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{i} ∈ E_{∞}(𝕄_{n}) and X_{n} ⊗ v^{i} ∈ E_{∞}(Γ_{n}); for n ≥ 2, pick elements X_{n,p−1} ∈ H^{2(n+p)−3}(𝕄_{n}), Z_{n,p−1} ∈ H^{2(n+p)−2}(𝕄_{n}) and Y_{n,p−1} ∈ H^{2(n+p)−2}(Γ_{n}) which represent respectively B_{n} ⊗ v^{p} ∈ E_{∞}(𝕄_{n}), x_{3}x_{4} … x_{2n−1}x_{2n} ⊗ v^{p} ∈ E_{∞}(𝕄_{n}) and A_{n} ⊗ v^{p} ∈ E_{∞}(Γ_{n}) (the existence of such elements follows from Propositions 2 and 8). Especially, define Y_{1,p−1} by

Y_{1,p−1} = N_{Ker x_{2}→Γ_{1}}(w)

with N the Evens norm map (note that Ker x_{2} ≅ C_{p^{2}} × C_{p} ⊂ Γ_{1}, so, by Künneth formula, w can be considered as an element of H^{2}(Ker x_{2})).

We now define the following subgroups of Γ_{n}:

𝕄_{n} = Ker (x − x_{1}),
Γ′_{n−1} = Ker x_{2n} ≅ Γ_{n−1} × C_{p},
𝕄′_{n−1} = Ker x_{2n} ∩ Ker (x − x_{1}) (so 𝕄′_{n−1} ≅ 𝕄_{n−1} × C_{p} for n > 1),
‘Γ_{n−1} = Ker x_{2} ∩ Ker (x − x_{1}) ≅ Γ_{n−1},
′Γ_{n−2} = Ker x_{2} ∩ Ker x_{2n} ∩ Ker (x − x_{1}) ≅ Γ_{n−2} × C_{p} (for n ≥ 2),

with the convention that Γ_{0} = C_{p^{2}}. Therefore ′Γ_{n−2} = 𝕄′_{n−1} ∩ ‘Γ_{n−1} and ′Γ_{0} = C_{p^{2}} × C_{p}. If K is one of the above subgroups, K contains Z as a central subgroup and we have the central extension

(K)  1 → Z → K → K/Z → 1.

For convenience, we also define the elements Y_{0,i} ∈ H^{2i}(Γ_{0}), 1 ≤ i ≤ p − 1, by Y_{0,i} = w^{i}. With some abuse of notation, by Künneth formula, the Y_{n−1,i} (resp. X_{n−1,i}, Z_{n−1,i})’s are considered as elements of H^{∗}(Γ′_{n−1}), H^{∗}(′Γ_{n−1}) and H^{∗}(‘Γ_{n−1}) (resp. H^{∗}(𝕄′_{n−1})). We have

Lemma 7. For n ≥ 1 and 1 ≤ i ≤ p − 1, 1 ≤ j ≤ p − 2,

(i) Res_{𝕄_{n}}^{Γ_{n}}(Y_{n,j}) + x_{2}X_{n,j} ∈ Im Inf_{𝕄_{n}}^{W}; if n > 1 then Res_{𝕄_{n}}^{Γ_{n}}(Y_{n,p−1}) + x_{2}X_{n,p−1} − Z_{n,p−1} ∈ Im Inf_{𝕄_{n}}^{W};

(ii) if n > 1 then Res_{‘Γ_{n−1}}^{𝕄_{n}}(X_{n,i}) − xY_{n−1,i} ∈ Im Inf_{‘Γ_{n−1}}^{‘Γ_{n−1}/Z};

(iii) Res_{Γ′_{n−1}}^{Γ_{n}}(Y_{n,j}) belongs to Im Inf_{Γ′_{n−1}}^{Γ′_{n−1}/Z}; (iv) xY_{n,i} ∉ (y); furthermore, there exists no element ξ ∈ H^{∗}(Γ_{n}) satisfying

xY_{n,i} = yξ mod Im Inf_{Γ_{n}}^{V}.

Proof. (i) and (ii) follow by considering the restriction in spectral sequences and by the structures of E_{2p+1}(𝕄_{n}) and E_{2p+1}(Γ_{n}) given in Theorem 1 (ii) and Proposition 2. (iii) Set T_{n,i} =Res_{Γ′_{n−1}}^{Γ_{n}}(Y_{n,i}). Since y_{2n−1}Y_{n,i} ∈Im Inf_{Γ_{n}}^{V}, it follows that y_{2n−1}T_{n,i} =

Res_{Γ′_{n−1}}^{Γ_{n}}(y_{2n−1}Y_{n,i}) belongs to Im Inf_{Γ′_{n−1}}^{Γ′_{n−1}/Z}. As H^{∗}(Γ′_{n−1}) = H^{∗}(Γ_{n−1})⊗E[x_{2n−1}]⊗

𝔽_{p}[y_{2n−1}], T_{n,i} also belongs to Im Inf_{Γ′_{n−1}}^{Γ′_{n−1}/Z}.

(iv) Assume that there exists ξ ∈ H^{2n+1}(Γ_{n}) such that yξ = xY_{n,1} mod Im Inf_{Γ_{n}}^{V}. By Proposition 3, ξ ∈ Im Inf_{Γ_{n}}^{V}. So yξ ∈ Im Inf_{Γ_{n}}^{V}. Hence xY_{n,1} ∈ Im Inf_{Γ_{n}}^{V}, a contradiction.

Assume inductively that (iv) holds for i − 1. For i ≥ 2, we will prove in Lemmas 10, 11, 16 and 17 that tr_{Γ_{n+1}}^{Γ′_{n}}(Y_{n,i}) = λ_{i}Y_{n+1,i−1} mod Im Inf_{Γ_{n+1}}^{V} with 0 ≠ λ_{i} ∈ 𝔽_{p}. Let ϕ be the element of Im Inf_{Γ_{n+1}}^{V} satisfying tr_{Γ_{n+1}}^{Γ′_{n}}(Y_{n,i}) = λ_{i}Y_{n+1,i−1} + ϕ. Suppose that xY_{n,i} + η = yξ, with ξ ∈ H^{∗}(Γ_{n}) and η ∈ Im Inf_{Γ_{n}}^{V}. So

λ_{i}xY_{n+1,i−1} + xϕ = xtr_{Γ_{n+1}}^{Γ′_{n}}(Y_{n,i})
= tr_{Γ_{n+1}}^{Γ′_{n}}(xY_{n,i} + η) since tr_{Γ_{n+1}}^{Γ′_{n}}(η) = 0
= tr_{Γ_{n+1}}^{Γ′_{n}}(yξ)
= ytr_{Γ_{n+1}}^{Γ′_{n}}(ξ).

Hence λ_{i}xY_{n+1,i−1} = ytr_{Γ_{n+1}}^{Γ′_{n}}(ξ) − xϕ which contradicts the inductive hypothesis. (iv) is then proved.

The lemma follows. □

Further properties of X_{n,i} and Y_{n,i} are given by the following lemmas. The first one follows from Theorem 1, Proposition 2 and [14, Theorem 1.1].

Lemma 8. tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,i}) (resp. tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i})) represents an element of E_{∞}^{∗,2j}(𝕄_{n}) (resp. E_{∞}^{∗,2j}(Γ_{n})), with j < i. □

Lemma 9. For 2 ≤ i ≤ p − 1, tr_{Γ_{1}}^{Γ′_{0}}(Y_{0,i}) = λ_{i}Y_{1,i−1} mod Im Inf_{Γ_{1}}^{V}, with 0 ≠ λ_{i} ∈ 𝔽_{p}.

Proof. Consider the commutative diagram

H^{∗}(C_{p^{2}} × C_{p}) ⟶^{tr} H^{∗}(Γ_{1}) ^{Res}↓  ↓^{Res} H^{∗}(Z × C_{p}) ⟶_{tr} H^{∗}(𝔼) ,

we have Res_{𝔼}^{Γ_{1}}tr_{Γ_{1}}^{Γ′_{0}}(Y_{0,i}) = tr_{𝔼}^{Z×C_{p}}(v^{i}). Following [8] (see also [16]), tr_{𝔼}^{Z×C_{p}}(v^{i}) is a non-zero element of H^{∗}(𝔼)\Im Inf_{𝔼}^{𝔼/Z}. So, by Theorem 1, tr_{Γ_{1}}^{Γ′_{0}}(Y_{0,i}) represents an element of the form λ_{i}x_{1}x_{2} ⊗ v^{i−1} ∈ E_{∞}^{2,2(i−1)}(Γ_{1}), with 0 ≠ λ_{i} ∈ 𝔽_{p}. The lemma follows. □

In the following two lemmas, p is assumed to be greater than 3.

Lemma 10. For 2 ≤ i ≤ p−2, tr_{𝕄_{2}}^{𝕄′_{1}}(X_{1,i}) = λ_{i}X_{2,i−1} mod Im Inf_{𝕄_{2}}^{W} and tr_{Γ_{2}}^{Γ′_{1}}(Y_{1,i}) = λ_{i}Y_{2,i−1} mod Im Inf_{Γ_{2}}^{V}, with λ_{i} given in Lemma 9.

Proof. Set Z_{i} = Res_{‘Γ_{1}}^{𝕄_{2}}tr_{𝕄_{2}}^{𝕄′_{1}}(X_{1,i}). By the double coset formula and by Lemma 7 (ii), we have

Z_{i} = tr_{‘Γ_{1}}^{′Γ_{0}}Res_{′Γ_{0}}^{𝕄′_{1}}(X_{1,i}) = tr_{‘Γ_{1}}^{′Γ_{0}}(x_{1}Y_{0,i}) = x_{1}tr_{‘Γ_{1}}^{′Γ_{0}}(Y_{0,i}).

By Lemma 9, Z_{i} represents λ_{i}x_{1}x_{3}x_{4} ⊗ v^{i−1} ∈ E_{∞}^{3,2(i−1)}(‘Γ_{1}). By Lemma 8 and Proposition 2, this means that tr_{𝕄_{2}}^{𝕄′_{1}}(X_{1,i}) represents λ_{i}x_{1}x_{3}x_{4}⊗v^{i−1} ∈ E_{∞}^{3,2(i−1)}(𝕄_{2}). The first part of the lemma follows from the definition of X_{2,i−1}.

On the other hand, by setting Y_{i} = Res_{𝕄_{2}}^{Γ_{2}}tr_{Γ_{2}}^{Γ′_{1}}(Y_{1,i}), by the double coset formula, we have

Y_{i} = tr_{𝕄_{2}}^{𝕄′_{1}}Res_{𝕄′_{1}}^{Γ′_{1}}(Y_{1,i})
= tr_{𝕄_{2}}^{𝕄′_{1}}(−x_{2}X_{1,i}) by Lemma 7 (i)
= −x_{2}tr_{𝕄_{2}}^{𝕄′_{1}}(X_{1,i}).

As showed above, tr_{𝕄_{2}}^{𝕄′_{1}}(X_{1,i}) represents λ_{i}x_{1}x_{3}x_{4} ⊗ v^{i−1} ∈ E_{∞}^{3,2(i−1)}(𝕄_{2}), so Y_{i} represents λ_{i}x_{1}x_{2}x_{3}x_{4} ⊗ v^{i−1} ∈ E_{∞}^{4,2(i−1)}(𝕄_{2}). By Lemma 8 and Theorem 1 (ii), this means that tr_{Γ_{2}}^{Γ′_{1}}(Y_{1,i}) represents λ_{i}x_{1}x_{2}x_{3}x_{4} ⊗ v^{i−1} ∈ E_{∞}^{4,2(i−1)}(Γ_{2}). The last part follows from the definition of Y_{2,i−1}. The lemma is proved. □

In general, we have

Lemma 11. For 2 ≤ i ≤ p − 2 and n ≥ 2, tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,i}) = λ_{i}X_{n,i−1} mod Im Inf_{𝕄_{n}}^{W}

and tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i}) = λ_{i}Y_{n,i−1} mod Im Inf_{Γ_{n}}^{V}, with λ_{i} given in Lemma 9.

Proof. We argue by induction on n. The case n = 2 follows from the above lemma. Assume that the lemma holds for n − 1. Set Z_{i} =Res_{‘Γ_{n−1}}^{𝕄_{n}}tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,i}). By the double coset formula, we have

Z_{i} = tr_{‘Γ_{n−1}}^{′Γ_{n−2}}Res_{′Γ_{n−2}}^{𝕄′_{n−1}}(X_{n−1,i})
= tr_{‘Γ_{n−1}}^{′Γ_{n−2}}(xY_{n−2,i}) by Lemma 7 (ii)
= xtr_{‘Γ_{n−1}}^{′Γ_{n−2}}(Y_{n−2,i}).

By the inductive hypothesis, tr_{‘Γ_{n−1}}^{′Γ_{n−2}}(Y_{n−2,i}) = λ_{i}Y_{n−1,i−1} mod Im Inf_{‘Γ_{n−1}}^{‘Γ_{n−1}/Z}. So

Z_{i} and λ_{i}xY_{n−1,i−1} represent the same element of E_{∞}^{2n+1,2(i−1)}(‘Γ_{n−1}). The first part follows from Lemma 8 and Proposition 2.

Finally, by setting Y_{i} = Res_{𝕄_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i}), we have

Y_{i} = tr_{𝕄_{n}}^{𝕄′_{n−1}}Res_{𝕄′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,i})
= tr_{𝕄_{n}}^{𝕄′_{n−1}}(−x_{2}X_{n−1,i}) by Lemma 7 (i)
= −x_{2}tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,i}).

As shown above, tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,i}) = λ_{i}X_{n,i−1}mod Im Inf_{𝕄_{n}}^{W}. So Y_{i} and −λ_{i}x_{2}X_{n,i−1} represent the same element of E_{∞}^{2n,2(i−1)}(𝕄_{n}).The last part follows from Lemma 8 and Theorem 1 (ii). The lemma is proved. □

We now calculate tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,1}). In so doing, let us recall the determination of the transfer map on Bar cochain levels. Let L, K be subgroups of Γ_{n} with Z ⊂ L ⊂ K and let D = \{d\} be the set of cosets of L in K. For each d, specify \overline{d} a representative of d such that \overline{L} = 1 and \overline{d} \overline{d′} \overline{dd′} ^{−1} ∈ Z. The transfer map t̃r_{K}^{L} : C^{∗}(L) → C^{∗}(K) is determined in [20] as follows:

\begin{aligned}t̃r_{K}^{L}f(·_{L}) & = ∑_{d∈D} f(·_{K}), \\ t̃r_{K}^{L}f(ℓ_{1}, …, ℓ_{n}) & = ∑_{d∈D} f(\overline{d}ℓ_{1}\overline{dℓ_{1}} ^{−1}, …, \overline{dℓ_{1} … ℓ_{n−1}}ℓ_{n}\overline{dℓ_{1} … ℓ_{n−1}ℓ_{n}} ^{−1}) \end{aligned}

for f ∈ C^{∗}(L), ℓ_{i} ∈ K.

Some properties of t̃r_{K}^{L} were given in [14]. Note that, if L is a direct factor of K, then t̃r_{K}^{L} is the zero map. Furthermore, if M is also a subgroup of Γ_{n} containing Z, we can choose representatives of the cosets of M in KM, and those of K ∩ M in K so that the double coset formula

(5)  Res_{K}^{KM} t̃r_{KM}^{M} = t̃r_{K}^{K∩M}Res_{K∩M}^{M}

holds at the cochain level.

Since v ∈ E_{2}(Γ′_{n−1}) is transgressive, there exists a 2-cochain ṽ of Γ′_{n−1} satisfying ṽ|_{Z} = v, δṽ = βz_{n−1} (see e.g. [15], for a determination of such a cochain). It follows from [14, Lemma 1.4] that t̃r_{Γ_{n}}^{Γ′_{n−1}}(βz_{n−1}) = 0, hence δt̃r_{Γ_{n}}^{Γ′_{n−1}}(ṽ) = t̃r_{Γ_{n}}^{Γ′_{n−1}}(δṽ) = 0;

in other words, t̃r_{Γ_{n}}^{Γ′_{n−1}}(ṽ) is a 2-cocycle of Γ_{n}. Set \overline{v} = [t̃r_{Γ_{n}}^{Γ′_{n−1}}(ṽ)] ∈ H^{2}(Γ_{n}) and let ẽ, ã_{1}, …, ã_{2n} be elements of Γ_{n} satisfying ẽZ = e, ã_{i}Z = a_{i} (recall that e, a_{1}, …, a_{2n} was defined in Section 2 as a basis of V of which the dual is x, x_{1}, …, x_{2n}), we have Lemma 12. \overline{v} = −x_{2n−1}x_{2n}.

Proof. Write \overline{v} = ∑_{1≤i≤2n}μ_{i}xx_{i} + ∑_{1≤i<j≤2n}μ_{ij}x_{i}x_{j} + ∑_{1≤i≤2n}ν_{i}y_{i} with μ_{i}, μ_{ij}, ν_{i} ∈ 𝔽_{p}

(note that, in H^{2}(Γ_{n}), y = −(x_{1}x_{2} + … x_{2n−1}x_{2n})). Consider the double coset formula (5) with M = Γ′_{n−1} and KM = Γ_{n} (this means that ã_{2n} ∈ K). For K = ⟨ẽ, ã_{i}, ã_{j}, ã_{2n}⟩ with 1 ≤ i, j ≤ 2n − 2, as ã_{2n} commutes with every element of K ∩ M, we have t̃r_{K}^{K∩M} = 0, so μ_{i} = μ_{2n} = ν_{i} = ν_{2n} = μ_{ij} = μ_{i 2n} = 0. For K = ⟨ẽ, ã_{i}, a_{2n−1}, ã_{2n}⟩ with 1 ≤ i ≤ 2n − 2, we have K ≅ Γ_{1} × C_{p}, K ∩ M = C_{p^{2}} × C_{p}^{2} and Res_{K∩M}^{M}(ṽ) = w; by a direct verification, we can show that tr_{K}^{K∩M}(w) = y, therefore [t̃r_{K}^{K∩M}(w)] = y = −x_{2n−1}x_{2n}, so μ_{2n−1} = μ_{i 2n−1} = ν_{2n−1} = 0 and μ_{2n−1 2n} = −1. The lemma follows. □

Lemma 13. For n ≥ 1, tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,1}) = −X_{n}; hence Res_{𝔼_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,1}) = 0.

Proof. A cocycle representing Y_{n−1,1} can be chosen as follows. Since x_{1}x_{2} … x_{2n−3} x_{2n−2} · βz_{n−1} = 0 in H^{∗}(Γ′_{n−1}/Z), there exists a cochain f of Γ′_{n−1}/Z (considered as a cochain of Γ′_{n−1} via the inflation map on cochains) satisfying δf = x_{1}x_{2} … x_{2n−3} x_{2n−2} · βz_{n−1}. Furthermore, it follows from the definition of ṽ that δ(x_{1}x_{2} … x_{2n−3} x_{2n−2}·ṽ) = x_{1}x_{2} … x_{2n−3}x_{2n−2}·βz_{n−1}, hence δ(x_{1}x_{2} … x_{2n−3}x_{2n−2}·ṽ−f) = 0. It is clear that g = x_{1}x_{2} … x_{2n−3}x_{2n−2}·ṽ−f is a cocycle representing X_{n}⊗v ∈ E_{∞}(Γ′_{n−1}).

Hence Y_{n−1,1} − [g] ∈ Im Inf_{Γ_{n−1}}^{Γ′_{n−1}/Z} which implies that tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,1}) is represented

by t̃r_{Γ_{n}}^{Γ′_{n−1}}(g). By [14, Lemma 1.4], t̃r_{Γ_{n}}^{Γ′_{n−1}}(g) = x_{1}x_{2} … x_{2n−3}x_{2n−2} · t̃r_{Γ_{n}}^{Γ′_{n−1}}(ṽ). So

[t̃r_{Γ_{n}}^{Γ′_{n−1}}(g)] = x_{1}x_{2} … x_{2n−3}x_{2n−2} · \overline{v}. The lemma follows from Lemma 12. □

Arguing as in the above proof, we can also choose a cocycle representing X_{n−1} ⊗v^{p} (which is non-zero in E_{∞}(Γ′_{n−1}), by Theorem 1, Propositions 5 and 7), as follows. As v^{p} ∈ E_{2}(Γ′_{n−1}) is transgressive and d_{2p+1}(v^{p}) = P^{1}βz_{n−1}, there exists a cochain ṽ^{p} of Γ′_{n−1} such that ṽ^{p}|_{Z} = v^{p} and δṽ^{p} = P^{1}βz_{n−1}. Let h be a cochain of Γ′_{n−1}/Z satisfying δh = P^{1}βz_{n−1} · X_{n−1}. We have

Lemma 14. k = k_{n} = ṽ^{p}·x_{1}x_{2} … x_{2n−3}x_{2n−2}−h is a cocycle representing X_{n−1}⊗v^{p} and Res_{𝔼_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}([k]) = 0.

Proof. It follows from the definitions of ṽ^{p} and h that k is a cocycle representing X_{n−1} ⊗ v^{p}. Set X = Res_{𝔼_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}([k]), then X = [t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}Res_{𝔼_{n−1}×C_{p}}^{Γ′_{n−1}}(k)] by the double coset formula. Denote also by ṽ^{p} (resp. h) the restriction of the cochain ṽ^{p} (resp. h) to 𝔼_{n−1} × C_{p}. By [14, Lemma 1.4], t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(h) = 0, hence X = [t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(ṽ^{p} · x_{1}x_{2} … x_{2n−3}x_{2n−2})]. Note that, in H^{∗}(𝔼_{n−1} × C_{p}), X_{n−1} = X_{n−2}(x_{1}x_{2} + ⋯ + x_{2n−3}x_{2n−2}) and x_{1}x_{2} + ⋯ + x_{2n−3}x_{2n−2} = 0, so there exist cochains c of (𝔼_{n−1} × C_{p})/Z and b of 𝔼_{n−1} × C_{p} satisfying

δb = x_{1}x_{2} + ⋯ + x_{2n−3}x_{2n−2},
and  x_{1}x_{2} … x_{2n−3}x_{2n−2} = x_{1}x_{2} … x_{2n−5}x_{2n−4} · δb + δc.

Hence

\begin{aligned}ṽ^{p} · x_{1}x_{2} … x_{2n−3}x_{2n−2} & = ṽ^{p} · x_{1}x_{2} … x_{2n−5}x_{2n−4} · δb + ṽ^{p} · δc \\ & = −δṽ^{p} · x_{1}x_{2} … x_{2n−5}x_{2n−4} · b − δṽ^{p} · c mod Im δ. \end{aligned}

So X = −[t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(δṽ^{p} · x_{1}x_{2} … x_{2n−5}x_{2n−4} · b + δṽ^{p} · c)]. Following [14, Lemma 1.4], t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(δṽ^{p} · c) = 0 and t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(δb) = 0. This implies that t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(b) is a cocycle of 𝔼_{n} and

X = −P^{1}βz_{n−1} · X_{n−2} · [t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(b)].

Arguing as in the proof of Lemma 12, we can show that [t̃r_{𝔼_{n}}^{𝔼_{n−1}×C_{p}}(b)] = 0. Hence X = 0. The lemma follows. □

With some abuse of notation, we also denote by ṽ (resp. ṽ^{p}) the restriction of ṽ (resp. ṽ^{p}) to 𝕄′_{n−1}. So δ(ṽ) = βz′_{n−1} and δ(ṽ^{p}) = P^{1}βz′_{n−1} in C^{∗}(𝕄′_{n−1}). Let ũ be a 1-cochain of 𝕄′_{n−1} satisfying δ(ũ) = z′_{n−1}. It follows from the proof of Proposition 8 that there exists a cochain d of 𝕄′_{n−1}/Z such that

\begin{aligned}δd & = x_{3}x_{4} … x_{2n−3}x_{2n−2}(P^{1}βz′_{n−1} − x_{2}z′_{n−1}^{p} − y_{2}^{p−1}βz′_{n−1} + y_{2}^{p−1}x_{2}z′_{n−1}) \\ & = δ(x_{3}x_{4} … x_{2n−3}x_{2n−2}(ṽ^{p} + x_{2}z′_{n−1}^{p−1}ũ − y_{2}^{p−1}ṽ − y_{2}^{p−1}x_{2}ũ)). \end{aligned}

So, for n ≥ 3, q = x_{3}x_{4} … x_{2n−3}x_{2n−2}(ṽ^{p} + x_{2}z′_{n−1}^{p−1}ũ − y_{2}^{p−1}ṽ − y_{2}^{p−1}x_{2}ũ) − d is a cocycle of 𝕄′_{n−1} representing Z_{n−1,p−1}. We have

Lemma 15. For n ≥ 3, tr_{𝕄_{n}}^{𝕄′_{n−1}}(Z_{n−1,p−1}) ∈ Im Inf_{𝕄_{n}}^{W}.

Proof. It follows that tr_{𝕄_{n}}^{𝕄′_{n−1}}(Z_{n−1,p−1}) = [t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(q)]. By [14, Lemma 1.4],

tr_{𝕄_{n}}^{𝕄′_{n−1}}(Z_{n−1,p−1}) = x_{3}x_{4} … x_{2n−3}x_{2n−2}([t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ^{p})] + x_{2}z′_{n−1}^{p−1}[t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ũ)]

− y_{2}^{p−1}[t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ)] − y_{2}^{p−1}x_{2}[t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ũ)])

(note that t̃r_{𝕄_{n}}^{𝕄′_{n−1}} maps each of ṽ^{p}, ũ, ṽ to a cocycle). Since each of t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ^{p}),

t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ), t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ũ) is of degree ≤ 2p, it follows from the structure of E_{2p+1}(𝕄_{n}) that

[t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ)], [t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ũ)] and the cup-product of x_{3}x_{4} … x_{2n−3}x_{2n−2} with [t̃r_{𝕄_{n}}^{𝕄′_{n−1}}(ṽ^{p})] belong to Im Inf_{𝕄_{n}}^{W}. The lemma follows. □

Lemma 16. There exists 0 ≠ λ ∈ 𝔽_{p} such that tr_{Γ_{2}}^{Γ′_{1}}(Y_{1,p−1}) − λY_{2,p−2} ∈ Im Inf_{Γ_{2}}^{V}.

Proof. Set K = Ker x_{2} ∩ Ker (x − x_{4}) ⊂ Γ_{2} and X = tr_{Γ_{2}}^{Γ′_{1}}(Y_{1,p−1}). So K ≅ 𝕄 ×

C_{p}, K ∩ Γ′_{1} ≅ C_{p}^{3} and Res_{K}^{Γ_{2}}(X) = tr_{K}^{Γ′_{1}∩K}Res_{Γ′_{1}∩K}^{Γ′_{1}}(Y_{1,p−1}). As Res_{Γ′_{1}∩K}^{Γ′_{1}}(Y_{1,p−1}) =

v^{p} − vy_{1}^{p−1}, we have

Res_{K}^{Γ_{2}}(X) = tr_{K}^{Γ′_{1}∩K}(v^{p} − vy_{1}^{p−1}) = tr_{K}^{Γ′_{1}∩K}(v^{p}) − tr_{K}^{Γ′_{1}∩K}(vy_{1}^{p−1}).

A direct verification shows that tr_{K}^{Γ′_{1}∩K}(v) = y_{4}, so Res_{K}^{Γ_{2}}(X) = −y_{4}y_{1}^{p−1} ≠ 0. Hence X ≠ 0.

Suppose that X ∈ Im Inf_{Γ_{2}}^{V}. Since y_{4}X = 0, y_{4}X must belong to (z, η_{1}, η_{2}, ξ_{1}). Write

(6)  y_{4}X = az + bη_{1} + cη_{2} + μξ_{1}

with a, b, c ∈ H^{∗}(V) and μ ∈ 𝔽_{p}. Multiplying (6) by x_{1}x_{2}x_{3}x_{4} yields μξ_{1} ∈ (y, y_{4}). Hence μ = 0. Multiplying (6) by η_{2} yields y_{4}Xη_{2} ∈ (z, η_{1}). So, by [13, Lemma 2.4], Xη_{2} ∈ (z, η_{1}, X_{2}). Since Xη_{2} is of degree > 4, it follows that Xη_{2} ∈ (z, η_{1}). By [13, Lemma 2.14], X = ey mod(z, η_{1}) with e ∈ H^{2p−2}(V). Write

(7)  eyy_{4} = a_{1}z + b_{1}η_{1} + c_{1}η_{2}.

Multiplying (7) by η_{1}η_{2} yields

\begin{aligned}eyy_{4}η_{1}η_{2} & = a_{1}zη_{1}η_{2} \\ & = a_{1}yη_{1}η_{2} − a_{1}X_{2}ξ_{1}. \end{aligned}

So a_{1} ∈ (y, x_{1}, …, x_{4}). Therefore b_{1} ∈ (y, x_{i}x_{j}) and c_{1} = 0. By [13, Lemma 2.4], we have ey ∈ (z, η_{1}, X_{2}). Since ey is of degree > 4, it follows that ey ∈ (z, η_{1}). So X ∈ (z, η_{1}), hence X = 0 in H^{∗}(Γ_{2}), a contradiction. The lemma follows. □

Lemma 17. For n ≥ 3, tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,p−1}) = λX_{n,p−2} mod Im Inf_{𝕄_{n}}^{W} and

tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,p−1}) = λY_{n,p−2} mod Im Inf_{Γ_{n}}^{V}, with λ given in Lemma 16.

Proof. Consider the case n = 3. Set X =Res_{‘Γ_{2}}^{𝕄_{3}}tr_{𝕄_{3}}^{𝕄′_{2}}(X_{2,p−1}). By the double coset formula, we have

X = tr_{‘Γ_{2}}^{′Γ_{1}}Res_{′Γ_{1}}^{𝕄′_{2}}(X_{2,p−1})
= tr_{‘Γ_{2}}^{′Γ_{1}}(xY_{1,p−1}) by Lemma 7 (ii)
= xtr_{‘Γ_{2}}^{′Γ_{1}}(Y_{1,p−1}).

It follows from Lemma 16 that X and λxY_{2,p−2} represent the same element of E_{∞}^{5,2(p−2)}(‘Γ_{2}). By Lemma 8 and Proposition 2, it follows that tr_{𝕄_{3}}^{𝕄′_{2}}(X_{2,p−1}) = λX_{3,p−2} mod Im Inf_{𝕄_{3}}^{W}. Similarly, by setting Y = Res_{𝕄_{3}}^{Γ_{3}}tr_{Γ_{3}}^{Γ′_{2}}(Y_{2,p−1}), we have

Y = tr_{𝕄_{3}}^{𝕄′_{2}}Res_{𝕄′_{2}}^{Γ′_{2}}(Y_{2,p−1})
= tr_{𝕄_{3}}^{𝕄′_{2}}(−x_{2}X_{2,p−1} + Z_{2,p−1}) by Lemma 7 (i)
= −x_{2}tr_{𝕄_{3}}^{𝕄′_{2}}(X_{2,p−1}) + tr_{𝕄_{3}}^{𝕄′_{2}}(Z_{2,p−1}).

As shown above, tr_{𝕄_{3}}^{𝕄′_{2}}(X_{2,p−1}) = λX_{3,p−2}mod Im Inf_{𝕄_{3}}^{W}. So, by Lemma 15, Y and −λx_{2}X_{3,p−2} represent the same element of E_{∞}^{6,2(p−2)}(𝕄_{3}). By Lemma 8 and Proposition 2, it follows that tr_{Γ_{3}}^{Γ′_{2}}(Y_{2,p−1}) = λY_{3,p−2} mod Im Inf_{Γ_{3}}^{V}.

Assume that the lemma holds for n − 1. Set Z =Res_{‘Γ_{n−1}}^{𝕄_{n}}tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,p−1}). By the double coset formula, we have

Z = tr_{‘Γ_{n−1}}^{′Γ_{n−2}}Res_{′Γ_{n−2}}^{𝕄′_{n−1}}(X_{n−1,p−1})
= tr_{‘Γ_{n−1}}^{′Γ_{n−2}}(xY_{n−2,p−1}) by Lemma 7 (ii)
= xtr_{‘Γ_{n−1}}^{′Γ_{n−2}}(Y_{n−2,p−1}).

By the inductive hypothesis, tr_{‘Γ_{n−1}}^{′Γ_{n−2}}(Y_{n−2,p−1}) = λY_{n−1,p−2} mod Im Inf_{‘Γ_{n−1}}^{‘Γ_{n−1}/Z}. So Z and λxY_{n−1,p−2} represent the same element of E_{∞}^{2n−1,2(p−2)}(‘Γ_{n−1}). The first part follows from Lemma 8 and Proposition 2.

Finally, by setting Y = Res_{𝕄_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,p−1}), we have

Y = tr_{𝕄_{n}}^{𝕄′_{n−1}}Res_{𝕄′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,p−1})
= tr_{𝕄_{n}}^{𝕄′_{n−1}}(−x_{2}X_{n−1,p−1}) + tr_{𝕄_{n}}^{𝕄′_{n−1}}(Z_{n−1,p−1}) by Lemma 7 (i)
= −x_{2}tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,p−1}) + tr_{𝕄_{n}}^{𝕄′_{n−1}}(Z_{n−1,p−1}).

As shown above, tr_{𝕄_{n}}^{𝕄′_{n−1}}(X_{n−1,p−1}) = λX_{n,p−2}mod Im Inf_{𝕄_{n}}^{W}. So, by Lemma 15, Y and −λx_{2}X_{n,p−2} represent the same element of E_{∞}^{2n,2(p−2)}(𝕄_{n}). The last part follows from Lemma 8 and Theorem 1 (ii). The lemma is proved. □

Let

⋯ ⊃ F^{i}C^{∗}(Γ′_{n−1}) ⊃ F^{i+1}C^{∗}(Γ′_{n−1}) ⊃ …

be the filtration of C^{∗}(Γ′_{n−1}) introduced by Hochschild-Serre ([7]) corresponding to the central extension (Γ′_{n−1}). Let us recall that

F^{i}C^{∗}(Γ′_{n−1}) = \{ \begin{matrix}C^{∗}(Γ′_{n−1}) & for i ≤ 0, \\ ∑_{m=0}^{∞} F^{i}C^{m}(Γ′_{n−1}) & for i > 0, \end{matrix}

where F^{i}C^{m}(Γ′_{n−1}) = 0 for i > m; and for 0 < i ≤ m, F^{i}C^{m}(Γ′_{n−1}) is the group of all m-cochains f for which f(g_{1}, …, g_{m}) = 0 whenever m − i + 1 of the arguments g_{k} belong to Z. It is clear that the conjugation by a = a_{2n} on C^{∗}(Γ′_{n−1}) is compatible with the Hochschild-Serre filtration. We have then the induced conjugation on Hochschild-Serre spectral sequence {E_{r}(Γ′_{n−1})}. As the action of a on E_{2}^{∗,∗}(Γ′_{n−1}) satisfies ^{a}x_{k} = x_{k}, 1 ≤ k ≤ 2n − 1, and ^{a}v = v + y_{2n−1}, it follows from the structure of E_{2p+1}(Γ′_{n−1}) that Y_{n−1,i} and ^{a}Y_{n−1,i} represent the same element of E_{∞}(Γ′_{n−1}). Hence

(8)  Y_{n−1,i} −^{a} Y_{n−1,i} = ∑_{0<j<i} μ_{j}Y_{n−1,j}y_{2n−1}^{i−j}+ ∑_{0<j<i} ν_{j}Y_{n−1,j}y_{2n−1}^{i−j−1}xx_{2n−1}

mod Im Inf_{Γ′_{n−1}}^{Γ′_{n−1}/Z},

with μ_{j}, ν_{j} ∈ 𝔽_{p}. We have

Lemma 18. For n ≥ 2, Y_{n−1,1} −^{a} Y_{n−1,1} = 0.

Proof. Set K = Ker x_{2n−2} ∩Γ′_{n−1}. Since the transfer commutes with the conjugation

and Im Inf_{Γ_{n−1}}^{Γ′_{n−1}/Z} is invariant under the action of a, by Lemmas 9, 10, 11, 16 and 17, we have

Y_{n−1,1} −^{a} Y_{n−1,1} = tr_{Γ′_{n−1}}^{K}(Y_{n−2,2} −^{a} Y_{n−2,2})

up to a non-zero constant multiple. By Lemma 13, tr_{Γ′_{n−1}}^{K}(Y_{n−2,1}) = −X_{n−1}, hence

tr_{Γ′_{n−1}}^{K}(Y_{n−2,1}y_{2n−3}) = 0 and tr_{Γ′_{n−1}}^{K}(Y_{n−2,1}x_{2n−3}) = 0 in H^{∗}(Γ′_{n−1}). The lemma

follows from (8) and from the fact that tr_{Γ′_{n−1}}^{K}Inf_{K}^{K/Z} = 0. □

We have then

Lemma 19. For n ≥ 2 and 1 ≤ i ≤ p − 1, Y_{n−1,i} +^{a} Y_{n−1,i} + ⋯ +^{a^{p−1}} Y_{n−1,i} = 0;

hence Res_{Γ′_{n−1}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i}) = 0.

Proof. Since 1 + a + ⋯ + a^{p−1} = (1 − a)^{p−1}, we need prove that ^{(1−a)^{p−1}}Y_{n−1,i} = 0. For 1 ≤ k ≤ p − 1, by (8) and by Lemma 18, ^{(1−a)^{k}}Y_{n−1,k} = 0. Since

Res_{Γ′_{n−1}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i}) =^{(1−a)^{p−1}} Y_{n−1,i}, the lemma follows. □

For n ≥ 2 and 1 ≤ i ≤ p − 2, set κ_{n,i} = Res_{𝔼_{n}}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i+1}). The proof of the theorem is completed by the following Proposition 9. 0 ≠ κ_{n,i} ∈ Ess(𝔼_{n}) with 1 ≤ i < p − 2 for p > 3, and i = 1 for p = 3.

Proof. It follows from Proposition 1, Lemmas 7 (iv), 9, 11 and 17 that κ_{n,i} ≠ 0 in H^{∗}(𝔼_{n}). Let K be a maximal subgroup of 𝔼_{n}. K is then of the form 𝔼_{n−1} × C_{p}. Let

L be the central product of K and C_{p^{2}} = ∩^{2n}_{j=1} Ker x_{j}. It follows that L is a subgroup

of Γ_{n} containing K and L ≅ Γ_{n−1} × C_{p}. Therefore

Res_{K}^{𝔼_{n}}(κ_{n,i}) = Res_{K}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i+1})
= Res_{K}^{L}Res_{L}^{Γ_{n}}tr_{Γ_{n}}^{Γ′_{n−1}}(Y_{n−1,i+1}).

Hence, if Γ_{n} = Γ′_{n−1}L, it follows from the double coset formula that

Res_{K}^{𝔼_{n}}(κ_{n,i}) = Res_{K}^{L}tr_{L}^{L∩Γ′_{n−1}}Res_{L∩Γ′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,i+1}).

By Lemma 1, there exists a non-zero linear combination α of x_{1}, …, x_{2n} such that L = Ker α. Consider the following cases:

• α = x_{2n−1} + γ with γ a linear combination of x_{1}, …, x_{2n−2}, x_{2n}: it follows that Γ_{n} = Γ′_{n−1}L and L ∩ Γ′_{n−1} ≅ Γ_{n−1} is a direct factor of L. Hence tr_{L}^{L∩Γ′_{n−1}} is the zero map. We have then

Res_{K}^{𝔼_{n}}(κ_{n,i}) = Res_{K}^{L}tr_{L}^{L∩Γ′_{n−1}}Res_{L∩Γ′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,i+1})

= 0;

• α = μx_{2n} + γ with γ a non-zero linear combination of x_{1}, …, x_{2n−2} and μ ∈ 𝔽_{p}: it follows that L ∩ Γ′_{n−1} = H × ⟨ã_{2n−1}⟩ for a subgroup H of Γ_{n−1} with H ≅ Γ′_{n−2}.

If p > 3, it follows from the proof of Lemma 7(iii) that Res_{L∩Γ′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,i+1}) belongs

to the ideal generated by Im Inf_{L∩Γ′_{n−1}}^{(L∩Γ′_{n−1})/Z}; since Im Inf_{L∩Γ′_{n−1}}^{(L∩Γ′_{n−1})/Z} ⊂ Ker tr_{L}^{L∩Γ′_{n−1}},

it follows that Res_{K}^{𝔼_{n}}(κ_{n,i}) = 0. If p = 3, by Lemma 14, there exist ϕ ∈ H^{2}(L ∩

Γ′_{n−1}), ψ ∈ H^{1}(L ∩ Γ′_{n−1}) such that Res_{L∩Γ′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,2}) is a linear combination of

[k_{n−1}], Y_{n−2,1} · ϕ, Y_{n−2,1} · xψ

and an element of Im Inf_{L∩Γ′_{n−1}}^{(L∩Γ′_{n−1})/Z}; since ϕ, ψ belong to Im Inf_{L∩Γ′_{n−1}}^{(L∩Γ′_{n−1})/Z}, by Lemmas

13 and 14, it follows that Res_{K}^{L}tr_{L}^{L∩Γ′_{n−1}}Res_{L∩Γ′_{n−1}}^{Γ′_{n−1}}(Y_{n−1,2}) = 0, so Res_{K}^{𝔼_{n}}(κ_{n,1}) = 0;

Finally, the case α = x_{2n} follows from Lemma 19. The proposition is proved. □

Acknowledgments. Most of the results of this paper were obtained during a stay at the University of Essen and the ETH-Zentrum in Autumn 1997. I would like to thank Eckart Viehweg, Hélène Esnault and Urs Stammbach for making the visits possible. Many thanks to David John Green for valuable comments.

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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