Chasing Maximal Pro-p Galois Groups via 1-Cyclotomicity

Abstract

Let p be a prime. We prove that certain amalgamated free pro-p products of Demushkin groups with pro-p-cyclic amalgam cannot give rise to a 1-cyclotomic oriented pro-p group, and thus do not occur as maximal pro-p Galois groups of fields containing a root of 1 of order p. We show that other cohomological obstructions which are used to detect pro-p groups that are not maximal pro-p Galois groups—the quadraticity of $\mathbb {Z}/p\mathbb {Z}$-cohomology and the vanishing of Massey products—fail with the above pro-p groups. Finally, we prove that the Minač–Tân pro-p group cannot give rise to a 1-cyclotomic oriented pro-p group, and we conjecture that every 1-cyclotomic oriented pro-p group satisfy the strong n-Massey vanishing property for $n=3,4$.

1 Introduction

Let p be a prime number, and let $1+p\mathbb {Z}_p$ denote the pro-p group of principal units of the ring of p-adic integers $\mathbb {Z}_p$—namely, $1+p\mathbb {Z}_p=\{1+p\lambda \,\mid \,\lambda \in \mathbb {Z}_p\}$. An oriented pro-p group is a pair $(G,\theta )$ consisting of a pro-p group G and a morphism of pro-p groups $\theta :G\rightarrow 1+p\mathbb {Z}_p$, called an orientation of G (see [29]; oriented pro-p groups were introduced by Efrat [7], with the name “cyclotomic pro-p pairs”). An oriented pro-p group $(G,\theta )$ gives rise to the continuous G-module $\mathbb {Z}_p(\theta )$, which is equal to $\mathbb {Z}_p$ as an abelian pro-p group, and which is endowed with the continuous G-action defined by

$$\begin{aligned} g\cdot \lambda =\theta (g)\cdot \lambda \qquad \text {for all }g\in G\text { and }\lambda \in \mathbb {Z}_p(\theta ). \end{aligned}$$

An oriented pro-p group $(G,\theta )$ is said to be Kummerian if the following cohomological condition is satisfied: for every $n\ge 1$ the natural morphism

$$\begin{aligned} \textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta ))\longrightarrow \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}), \end{aligned}$$

(1.1)

induced by the epimorphism of continuous G-modules $\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )\twoheadrightarrow \mathbb {Z}/p$ is surjective (see [11])—here we consider $\mathbb {Z}/p$ as a trivial G-module. Moreover, the oriented pro-p group $(G,\theta )$ is said to be 1-cyclotomic if the above cohomological condition is satisfied also for every closed subgroup of G—namely, the natural morphism (1.1) is surjective also with H instead of G, and the restriction $\theta \vert _H:H\rightarrow 1+p\mathbb {Z}_p$ instead of $\theta$ for all closed subgroups H of G (in [25, 26] a 1-cyclotomic oriented pro-p group is called a “1-smooth” oriented pro-p group). This cohomological condition was considered first by J. Labute, who showed ante litteram that for every Demushkin group G there exists precisely one orientation which completes G into a Kummerian oriented pro-p group, namely the orientation induced by the dualizing module of G (see [13]).

In case of trivial orientations, 1-cyclotomicity translates into a purely group-theoretical statement. Namely, an oriented pro-p group $(G,\textbf{1})$—where $\textbf{1}:G\rightarrow 1+p\mathbb {Z}_p$ denotes the orientation which is constantly equal to 1—is 1-cyclotomic if, and only if, the abelianization of every closed subgroup of G is a free abelian pro-p group. Pro-p groups satisfying this group-theoretic condition are called absolutely torsion-free pro-p groups, and they were introduced by Würfel [36].

The main goal of this work is to produce new examples of pro-p groups which no orientations can turn into a 1-cyclotomic oriented pro-p group.

Theorem 1.1

Let G be a pro-p group with pro-p presentation

$$\begin{aligned} G =\left\langle \, x,y_1,\ldots ,y_{d_1},z_1,\ldots ,z_{d_2}\,\mid \, r_1=r_2=1 \,\right\rangle , \end{aligned}$$

(1.2)

where $d_1,d_2$ are two positive odd integers, and either:

• (1.1.a) $d_1+d_2\ge 4$ and

  $$\begin{aligned}\begin{aligned} r_1&=[x,y_1][y_2,y_3]\cdots [y_{d_1-1},y_{d_1}], \\ r_2&=[x,z_1][z_2,z_3]\cdots [z_{d_2-1},z_{d_2}]; \end{aligned} \end{aligned}$$

• (1.1.b) or p is odd and

  $$\begin{aligned}\begin{aligned} r_1&=y_1^p[y_1,x][y_2,y_3]\cdots [y_{d_1-1},y_{d_1}],\\ r_2&= z_1^p[z_1,x][z_2,z_3]\cdots [z_{d_2-1},z_{d_2}]. \end{aligned}\end{aligned}$$

Then there are no orientations $\theta :G\rightarrow 1+p\mathbb {Z}_p$ such that the oriented pro-p group $(G,\theta )$ is 1-cyclotomic.

It is worth underlining that the pro-p groups described in Theorem 1.1 are amalgamated free pro-p products of two Demushkin groups—the subgroup generated by $x,y_1,\ldots ,y_{d_1}$ and the subgroup generated by $x,z_1,\ldots ,z_{d_2}$—, with pro-p-cyclic amalgam, generated by x. Despite Demushkin groups and their free pro-p products are some of the (extremely few) examples of pro-p groups which are known to give rise to 1-cyclotomic oriented pro-p groups, the presence of a pro-p-cyclic amalgam is enough to lose 1-cyclotomicity.

Oriented pro-p groups satisfying 1-cyclotomicity have great prominence in Galois theory. Given a field $\mathbb {K}$, let ${\bar{\mathbb {K}}}_s$ and $\mathbb {K}(p)$ denote respectively the separable closure of $\mathbb {K}$, and the compositum of all finite Galois p-extensions of $\mathbb {K}$. The maximal pro-p Galois group of $\mathbb {K}$, denoted by $G_{\mathbb {K}}(p)$, is the maximal pro-p quotient of the absolute Galois group ${{\,\textrm{Gal}\,}}({\bar{\mathbb {K}}}_s/\mathbb {K})$ of ${\mathbb {K}}$, and it coincides with the Galois group of the Galois extension $\mathbb {K}(p)/\mathbb {K}$. Detecting maximal pro-p Galois groups among pro-p groups, are crucial problems in Galois theory. Already the pursuit of concrete examples of pro-p groups which do not occur as maximal pro-p Galois groups of fields is already considered a very remarkable challenge (see, e.g., [1, 3, 4, 24, 33]).

The maximal pro-p Galois group $G_{\mathbb {K}}(p)$ of a field $\mathbb {K}$ containing a root of 1 of order p gives rise to the oriented pro-p group $(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$, where

$$\begin{aligned} \theta _{\mathbb {K}}:G_{\mathbb {K}}(p)\longrightarrow 1+p\mathbb {Z}_p \end{aligned}$$

denotes the pro-p cyclotomic character (see Example 2.4 below). By Kummer theory, the oriented pro-p group $(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$ is 1-cyclotomic (see [13, p. 131] and [11, Sect. 4])—in case $p=2$ we need to assume further that $\sqrt{-1}\in \mathbb {K}$. Therefore, a pro-p group which cannot complete into a 1-cyclotomic oriented pro-p group does not occur as the maximal pro-p group of a field containing a root of 1 of order p—and hence neither as the absolute Galois group of any field (see, e.g., [24, Remark 3.3]). Hence, the following corollary may be deduced directly from Theorem 1.1.

Corollary 1.2

A pro-p group G as in Theorem 1.1 does not occur as the maximal pro-p Galois group of any field containing a root of 1 of order p (and also $\sqrt{-1}$ if $p=2$). Hence, G does not occur as the absolute Galois group of any field.

In the recent past, other cohomological properties have been used to study maximal pro-p Galois groups—and to find examples of pro-p groups which do not occur as maximal pro-p Galois groups. By the Norm Residue Theorem—proved by M. Rost and V. Voevodsky, with the contribution by Ch. Weibel, see [12, 34]—one knows that if $\mathbb {K}$ is a field containing a root of 1 of order p, the $\mathbb {Z}/p$ -cohomology algebra $\textbf{H}^\bullet (G_{\mathbb {K}}(p),\mathbb {Z}/p\mathbb {Z})$, endowed with the cup-product

$$\begin{aligned} \textvisiblespace \smallsmile \textvisiblespace :\textrm{H}^m(G_{\mathbb {K}}(p),\mathbb {Z}/p\mathbb {Z})\times \textrm{H}^n(G_{\mathbb {K}}(p),\mathbb {Z}/p\mathbb {Z})\longrightarrow \textrm{H}^{m+n}(G_{\mathbb {K}}(p),\mathbb {Z}/p\mathbb {Z}), \end{aligned}$$

is quadratic, i.e., its ring structure is completely determined by the 1st and the 2nd cohomology groups (see, e.g., [22, Sect. 2]). Moreover, it was shown by E. Matzri that if $\mathbb {K}$ is a field containing a root of 1 of order p, then $G_{\mathbb {K}}(p)$ satisfies the triple Massey vanishing property (see [9] and references therein)—for an overview on Massey products in Galois cohomology see [19]. These two cohomological properties were used to find examples of pro-p groups which do not occur as maximal pro-p Galois groups of fields containing a root of 1 of order p, for example in [4, Sect. 8] and in [19, Sect. 7].

We prove that the pro-p groups described in Theorem 1.1 cannot be ruled out as maximal pro-p Galois groups employing the above two cohomological obstructions.

Proposition 1.3

Let G be a pro-p group as in Theorem 1.1.

• (i) The $\mathbb {Z}/p$-cohomology algebra $\textbf{H}^\bullet (G,\mathbb {Z}/p\mathbb {Z})$ is quadratic.

• (ii) The pro-p group G satisfies the cyclic p-Massey vanishing property—namely, the p-fold Massey product

  $$\begin{aligned} \langle \underbrace{\alpha ,\ldots ,\alpha }_{p\text { times}}\rangle \end{aligned}$$

  contains 0 for every $\alpha \in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$.

• (iii.a) If G is as in (1.1.a), then G satisfies the 3- and the strong 4-Massey vanishing property.

• (iii.b) If G is as in (1.1.b) and $p>3$ then G satisfies the 3- and the strong 4-Massey vanishing property.

(We recall the basic notions on Massey products in Galois cohomology in Sect. 6.1 below.) Hence, Corollary 1.2 provides brand new examples of pro-p groups which do not occur as maximal pro-p Galois groups of fields containing a root of 1 of order p, and as absolute Galois groups. Moreover, we remark that the relations which define the pro-p groups described in Theorem 1.1 are rather “elementary”—just elementary commutators of generator times, possibly, the p-power of a generator—, unlike the examples provided in [1, 4, 19, 24], where the relations involve higher commutators.

Finally, we focus on the Minač–Tân pro-p group, i.e., the pro-p group G with pro-p presentation

$$\begin{aligned} G=\langle \,x_1,\ldots ,x_5\,\mid \,[[x_1,x_2],x_3][x_4,x_5]=1\,\rangle . \end{aligned}$$

In [19, Sect. 7], Minač and Tân showed that G does not satisfy the 3-Massey vanishing property, and thus it does not occur as the maximal pro-p Galois group of any field containing a root of 1 of order p. We prove that G cannot complete into a 1-cyclotomic oriented pro-p group.

Theorem 1.4

Let p be an odd prime. Then there are no orientations turning the Minač–Tân pro-p group into a 1-cyclotomic oriented pro-p group.

Theorem 1.4 has been proved independently by I. Snopce and P. Zalesskiĭ (unpublished). Theorem 1.4 provides a negative answer to the question posed in [29, Remark 3.7]—namely, the Minač–Tân pro-p group may be ruled out as a maximal pro-p Galois group of a field containing a root of 1 of order p (and thus as an absolute Galois group) in a “Massey-free” way.

Altogether, 1-cyclotomicity of oriented pro-p groups provides a rather powerful tool studying maximal pro-p Galois groups, and it succeeds in detecting pro-p groups which are not maximal pro-p Galois groups when other methods fail, as underlined above. We believe that further investigations in this direction will lead to new obstructions for the realization of pro-p groups as maximal pro-p Galois group.

Actually, Theorem 1.4, and the main result in [33] (see in particular [33, p. 1907]), may lead to the suspect that 1-cyclotomicity is a more restrictive condition in comparison with the vanishing of Massey products. Thus, we formulate the following conjecture.

Conjecture 1.5

Let $(G,\theta )$ be an oriented pro-p group, such that ${{\,\textrm{Im}\,}}(\theta )\subseteq 1+4\mathbb {Z}_2$ if $p=2$. If $(G,\theta )$ is 1-cyclotomic, then the pro-p group G satisfies the 3-Massey vanishing property; if moreover G is finitely generated, then G satisfies the strong n-Massey vanishing property for every $n\ge 3$.

After the publication on the arXiv of an earlier version of this paper, A. Merkurjev and F. Scavia proved the first statement of Conjecture 1.5—see [16, Theorem 1.3]—; while, on the other hand, there are 1-cyclotomic oriented pro-2 groups $(G,\theta )$ such that ${{\,\textrm{Im}\,}}(\theta )\subseteq 1+4\mathbb {Z}_2$, where G is not finitely generated and does not satisfy the strong 4-Massey vanishing property—see [15, Theorem 1.6]. In particular, [16, Theorem 1.3] implies Theorem 1.4 (see also [16, Remark 6.3]).

2 Oriented Pro-p Groups and Cohomology

2.1 Notation and Preliminaries

Throughout the paper, every subgroup of a pro-p group is tacitly assumed to be closed with respect to the pro-p topology. Therefore, sets of generators of pro-p groups, and presentations, are to be intended in the topological sense.

Given a pro-p group G, we denote the closed commutator subgroup of G by $G'$—namely, $G'$ is the closed normal subgroup generated by commutators

$$\begin{aligned}{}[h,g]= h^{-1}\cdot h^g=h^{-1}\cdot g^{-1}hg,\qquad g,h\in G. \end{aligned}$$

The Frattini subgroup of G is denoted by $\Phi (G)$—namely, $\Phi (G)$ is the closed normal subgroup generated by $G'$ and by p-powers $g^p$, $g\in G$ (cf., e.g., [5, Proposition 1.13]). A minimal generating set of G gives rise to a basis of the $\mathbb {Z}/p\mathbb {Z}$-vector space $G/\Phi (G)$, and conversely (cf., e.g., [5, Proposition 1.9]).

Finally, we denote the abelianization $G/G'$ of G by $G^{{{\,\textrm{ab}\,}}}$. Throughout the paper, we will make use of the following straightforward fact.

Fact 2.1

Let G be a finitely generated pro-p group. Then a subset $\{x_1,\ldots ,x_d\}$ of G is a minimal generating set of G if, and only if, the subset $\{x_1G',\ldots ,x_dG'\}$ of $G^{{{\,\textrm{ab}\,}}}$ is a minimal generating set of the abelian pro-p group $G^{{{\,\textrm{ab}\,}}}$.

2.2 Oriented Pro-p Groups

Let G be a pro-p group. An orientation $\theta :G\rightarrow 1+p\mathbb {Z}_p$ is said to be torsion-free if p is odd, or if $p=2$ and ${{\,\textrm{Im}\,}}(\theta )\subseteq 1+4\mathbb {Z}_2$. Observe that one may have an oriented pro-p group $(G,\theta )$ where G has non-trivial torsion and $\theta$ torsion-free (e.g., if $G\simeq \mathbb {Z}/p$ and ${{\,\textrm{Im}\,}}(\theta )=\{1\}$).

A morphism of oriented pro-p groups $(G_1,\theta _1)\rightarrow (G_2,\theta _2)$, is a homomorphism of pro-p groups $\phi :G_1\rightarrow G_2$ such that $\theta _1=\theta _2\circ \phi$ (cf. [29, Sect. 3, p. 1888]).

Within the family of oriented pro-p groups one has the following constructions. Let $(G,\theta )$ be an oriented pro-p group.

• (a) If N is a normal subgroup of G contained in ${{\,\textrm{Ker}\,}}(\theta )$, one has the oriented pro-p group $(G/N,\theta _{/N})$, where $\theta _{/N}:G/N\rightarrow 1+p\mathbb {Z}_p$ is the orientation such that $\theta _{/N}\circ \pi =\theta$, with $\pi :G\rightarrow G/N$ the canonical projection.

• (b) If A is an abelian pro-p group (written multiplicatively), one has the oriented pro-p group $A\rtimes (G,\theta )=(A\rtimes G,{{\tilde{\theta }}})$, with action given by $gag^{-1}=a^{\theta (g)}$ for every $g\in G$, $a\in A$, where the orientation ${{\tilde{\theta }}}:A\rtimes G\rightarrow 1+p\mathbb {Z}_p$ is the composition of the canonical projection $A\rtimes G\rightarrow G$ with $\theta$.

2.3 Kummerianity and 1-Cyclotomicity

Let $(G,\theta )$ be an oriented pro-p group. Observe that the G-action on the G-module $\mathbb {Z}_p(\theta )/p\mathbb {Z}_p(\theta )$ is trivial, as $\theta (g)\equiv 1\bmod p$ for all $g\in G$. Thus, $\mathbb {Z}_p(\theta )/p\mathbb {Z}_p(\theta )$ is isomorphic to $\mathbb {Z}/p$ as a trivial G-module.

An oriented pro-p group $(G,\theta )$ comes endowed with the distinguished subgroup

$$\begin{aligned} K_\theta (G)=\left\langle \,{}^{g}h\cdot h^{-\theta (g)}\,\mid \,g\in G,\,h\in {{\,\textrm{Ker}\,}}(\theta )\,\right\rangle \end{aligned}$$

(cf. [11, Sect. 3]). The subgroup $K_\theta (G)$ is normal in G, and it is contained in both ${{\,\textrm{Ker}\,}}(\theta )$ and $\Phi (G)$. On the other hand, $K_\theta (G)\supseteq {{\,\textrm{Ker}\,}}(\theta )'$, so that ${{\,\textrm{Ker}\,}}(\theta )/K_\theta (G)$ is an abelian pro-p group. Moreover, if $\theta$ is a torsion-free orientation, $G/{{\,\textrm{Ker}\,}}(\theta )\simeq {{\,\textrm{Im}\,}}(\theta )$ is torsion-free, and thus either trivial or isomorphic to $\mathbb {Z}_p$. Hence, the epimorphism $G\twoheadrightarrow G/{{\,\textrm{Ker}\,}}(\theta )$ splits, and since $ghg^{-1}\equiv h^{\theta (g)}\bmod K_\theta (G)$ for every $g\in G$ and $h\in {{\,\textrm{Ker}\,}}(\theta )$, one concludes that

$$\begin{aligned} \left( G/K_{\theta }(G),\theta _{/K_{\theta }(G)}\right) \simeq \dfrac{{{\,\textrm{Ker}\,}}(\theta )}{K_\theta (G)}\rtimes \left( G/{{\,\textrm{Ker}\,}}(\theta ),\theta _{/{{\,\textrm{Ker}\,}}(\theta )}\right) \end{aligned}$$

(cf., e.g., [30, Sect. 2.2, Eq. (2.1)]).

One has the following result relating the subgroup $K_\theta (G)$ and the surjectivity of the maps (1.1) (cf. [11, Theorem 7.1], see also [30, Proposition 2.6]).

Proposition 2.2

Let $(G,\theta )$ be an oriented pro-p group with $\theta$ a torsion-free orientation. The following are equivalent.

• (i) The natural map

  $$\begin{aligned} \textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta ))\longrightarrow \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}), \end{aligned}$$

  is surjective for every positive integer n.

• (ii) The quotient ${{\,\textrm{Ker}\,}}(\theta )/K_\theta (G)$ is a free abelian pro-p group.

If an oriented pro-p group $(G,\theta )$ with torsion-free orientation satisfies the above two equivalent properties, then it is said to be Kummerian. Moreover, $(G,\theta )$ is said to be 1-cyclotomic if $(H,\theta \vert _H)$ is Kummerian for every subgroup $H\subseteq G$.

Remark 2.3

The original definition of 1-cyclotomic oriented pro-p group requires only that for every open subgroup U of G, the oriented pro-p group $(U,\theta \vert _U)$ is Kummerian (cf. [29, Sect. 1]). By a continuity argument, this is enough to imply that the oriented pro-p group $(H,\theta \vert _H)$ is Kummerian also for every subgroup $H\subseteq G$ (cf. [29, Corollary 3.2]).

If $(G,\textbf{1})$ is an oriented pro-p group with $\textbf{1}:G\rightarrow 1+p\mathbb {Z}_p$ the orientation constantly equal to 1, then $K_{\textbf{1}}(G)=G'$, and by Proposition 2.2$(G,\theta )$ is Kummerian if, and only if, $G/G'={{\,\textrm{Ker}\,}}(\textbf{1})/K_{\textbf{1}}(G)$ is a free abelian pro-p group (cf. [11, Example 3.5–(1)]). Hence, $(G,\textbf{1})$ is 1-cyclotomic if, and only if, $H/H'$ is a free abelian pro-p group for every subgroup $H\subseteq G$, i.e., G is absolutely torsion-free (cf. [25, Remark 2.3]).

2.4 Examples

Example 2.4

Let $\mathbb {K}$ be a field containing a root of 1 of order p, and also $\sqrt{-1}$ if $p=2$. Then the pro-p cyclotomic character $\theta _{\mathbb {K}}$ of $G_{K}(p)$—induced by the action of $G_{\mathbb {K}}(p)$ on the roots of 1 of p-power order contained in $\mathbb {K}(p)$—has image contained in $1+p\mathbb {Z}_p$. Observe that ${{\,\textrm{Im}\,}}(\theta _{\mathbb {K}})=1+p^f\mathbb {Z}_p$, where $f\in \mathbb {N}\cup \{\infty \}$ is maximal such that $\mathbb {K}$ contains a root of 1 of order $p^f$ (if $f=\infty$, we set $p^\infty =0$). In particular, $\theta _{\mathbb {K}}$ is a torsion-free orientation. The module $\mathbb {Z}_p(\theta _{\mathbb {K}})$ is called the 1st Tate twist of $\mathbb {Z}_p$ (cf., e.g., [20, Definition 7.3.6]).

For the convenience of the reader, here we recall J. Labute’s argument to show that the oriented pro-p group $(G_{\mathbb {K}}(p),\theta _{\mathbb {K}})$ is Kummerian—and thus also 1-cyclotomic, as every subgroup $H\subseteq G_{\mathbb {K}}(p)$ is the maximal pro-p Galois group of an extension of $\mathbb {K}$, with pro-p cyclotomic character $\theta _{\mathbb {K}}\vert _H$—as it is presented in [13, p. 131] (where the module $\mathbb {Z}_p(\theta _{\mathbb {K}})$ is denoted by $I=I(\chi ')$). For every $n\ge 1$ one has an isomorphism of continuous $G_{\mathbb {K}}(p)$-modules

$$\begin{aligned} \mathbb {Z}_p(\theta _{\mathbb {K}})/p^n\mathbb {Z}_p(\theta _\mathbb {K})\simeq \mu _{p^n}=\left\{ \,\zeta \in \mathbb {K}(p)\,\mid \,\zeta ^{p^n}=1\,\right\} . \end{aligned}$$

Let $\mathbb {K}^\times$ and $\mathbb {K}(p)^\times$ denote the multiplicative groups of units of $\mathbb {K}$ and $\mathbb {K}(p)$ respectively. By Hilbert 90, the short exact sequence of continuous $G_{\mathbb {K}}(p)$-modules

(2.1)

induces a commutative diagram

where the left-side vertical arrow $\pi _n$ and the central vertical arrow are induced by the $p^{n-1}$th power map $\textvisiblespace ^{p^n}:\mathbb {K}(p)^\times \rightarrow \mathbb {K}(p)^\times$, and the right-side vertical arrow is induced by the epimorphism of continuous $G_{\mathbb {K}}(p)$-modules $\mathbb {Z}_p(\theta _{\mathbb {K}})/p^n\mathbb {Z}_p(\theta _{\mathbb {K}})\twoheadrightarrow \mathbb {Z}/p\mathbb {Z}$. Since the map $\pi _n$ is surjective, also the other vertical arrows are surjective.

Example 2.5

Let G be a free pro-p group. Then the oriented pro-p group $(G,\theta )$ is 1-cyclotomic for any orientation $\theta :G\rightarrow 1+p\mathbb {Z}_p$ (cf. [29, Sect. 2.2]).

Example 2.6

Let G be an infinite Demushkin group (cf., e.g., [20, Definition 3.9.9]). By [13, Theorem 4], G comes endowed with a canonical orientation $\chi :G\rightarrow 1+p\mathbb {Z}_p$ which is the only one completing G into a 1-cyclotomic oriented pro-p group. In particular, if $d=\dim (\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}))$ is even (which is always the case if $p\ne 2$), then G has a presentation

$$\begin{aligned} G=\left\langle \,x_1,\ldots ,x_d\,\mid \,x_1^{p^f}[x_1,x_2]\cdots [x_{d-1},x_d]=1\,\right\rangle , \end{aligned}$$

with $f\ge 1$ ($f\ge 2$ if $p=2$). In this case $\chi (x_2)=(1-p^f)^{-1}$ and $\chi (x_i)=1$ for $i\ne 2$.

Example 2.7

Let $(G,\theta )$ be an oriented pro-p group, with $\theta$ a torsion-free orientation. The oriented pro-p group $(G,\theta )$ is said to be $\theta$ -abelian if the subgroup $K_\theta (G)$ is trivial and if ${{\,\textrm{Ker}\,}}(\theta )$ is a free abelian pro-p group—in this case G is a free abelian-by-cyclic pro-p group, i.e.,

$$\begin{aligned} G\simeq {{\,\textrm{Ker}\,}}(\theta )\rtimes \dfrac{G}{{{\,\textrm{Ker}\,}}(\theta )} \end{aligned}$$

(cf. [30, Remark 2.2]). In other words, G has a presentation

$$\begin{aligned} G=\left\langle \,x_0,x_i\,\mid \,i\in I,\, x_i^{x_0}=x_i^{\theta (x_0)^{-1}},[x_i,x_j]=1\,\forall \,i,j\in I\,\right\rangle , \end{aligned}$$

for some set of indices I, and $\theta (x_i)=1$ for all $i\in I$ (cf. [22, Proposition 3.4]). A $\theta$-abelian oriented pro-p group $(G,\theta )$ is Kummerian by Proposition 2.2, as by definition $K_\theta (G)$ is trivial and ${{\,\textrm{Ker}\,}}(\theta )$ is a free abelian pro-p group. Moreover, if H is a subgroup of G, then one has

$$\begin{aligned} H\simeq (H\cap {{\,\textrm{Ker}\,}}(\theta ))\rtimes \frac{H}{{{\,\textrm{Ker}\,}}(\theta \vert _H)} \end{aligned}$$

(cf. [30, Remark 2.4]), so that the oriented pro-p group $(H,\theta \vert _H)$ is $\theta \vert _H$-abelian, and thus Kummerian, and consequently $(G,\theta )$ is 1-cyclotomic.

One has the following result to check whether an oriented pro-p group is Kummerian (cf. [30, Propositions 2.6, 3.6]).

Proposition 2.8

Let $(G,\theta )$ be an oriented pro-p group, with $\theta$ a torsion-free orientation. Then $(G,\theta )$ is Kummerian if, and only if, there exists a normal subgroup N of G such that $N\subseteq {{\,\textrm{Ker}\,}}(\theta )\cap \Phi (G)$, and the quotient $(G/N,\theta _{/N})$, is a $\theta _{/N}$-abelian oriented pro-p group. If such a normal subgroup N exists, then $N=K_\theta (G)$.

2.5 Kummerianity and 1-Cocyles

Let $(G,\theta )$ be an oriented pro-p group. Recall that for $n\in \mathbb {N}\cup \{\infty \}$, a 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ is a continuous map satisfying

$$\begin{aligned} c(gh)=c(g)+{\overline{\theta (g)}}c(h)\qquad \text {for every }g,h\in G, \end{aligned}$$

(2.2)

where $\overline{\theta (g)}$ denotes the image of $\theta (g)$ modulo $p^n$. From (2.2) one deduces

$$\begin{aligned} c([g,h])=\overline{\theta (gh)^{-1}}\left( c(g)(1-\overline{\theta (h)})-c(h)(1-\overline{\theta (g)})\right) . \end{aligned}$$

(2.3)

For $n\in \mathbb {N}\cup \{\infty \}$, every element of $\textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta ))$ is represented by a 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$. The following result is due to J. Labute (cf. [13, Proposition 6]).

Lemma 2.9

Let $(G,\theta )$ be a finitely generated oriented pro-p group with torsion-free orientation, and let $\mathcal {X}=\{x_1,\ldots ,x_d\}$ be a minimal generating set of G. The following are equivalent.

• (i) $(G,\theta )$ is Kummerian.

• (ii) For all $n\in \mathbb {N}\cup \{\infty \}$ and for any sequence $\lambda _1,\ldots ,\lambda _d$ of elements of $\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ there exists a continuous 1-cocycle $G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ satisfying $c(x_i)=\lambda _i$ for all $i=1,\ldots ,d$.

Proposition 2.10

Let G be a finitely generated pro-p group, and let $(G,\theta )$ be a Kummerian oriented pro-p group with torsion-free orientation. If N is a normal subgroup of G such that $N\subseteq {{\,\textrm{Ker}\,}}(\theta )$ and the restriction map

$$\begin{aligned} \textrm{res}_{G,N}^1:\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})\longrightarrow \textrm{H}^1(N,\mathbb {Z}/p\mathbb {Z})^G \end{aligned}$$

is surjective, then also $(G/N,\theta _{/N})$ is Kummerian.

In order to prove Proposition 2.10 we need the following fact, whose proof—rather straightforward—is left to the reader.

Fact 2.11

Let G be a finitely generated pro-p group, and let $(G,\theta )$ be an oriented pro-p group with torsion-free orientation.

• (i) If $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ is a continuous 1-cocycle, with $n\in \mathbb {N}\cup \{\infty \}$, then $c^{-1}(0)\cap {{\,\textrm{Ker}\,}}(\theta )$ is a normal subgroup of G.

• (ii) Let $N\subseteq G$ be a normal subgroup satisfying $N\subseteq {{\,\textrm{Ker}\,}}(\theta )$, with canonical projection $\pi :G\rightarrow G/N$. For $n\in \mathbb {N}\cup \{\infty \}$ one has the following:

  • (a) a continuous 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ satisfying $c\vert _N\equiv 0$ induces a continuous 1-cocycle ${\bar{c}} :G/N\rightarrow \mathbb {Z}_p(\theta _{/N})/p^n\mathbb {Z}_p(\theta _{/N})$ such that $c={\bar{c}}\circ \pi$;

  • (b) a continuous 1-cocycle ${\bar{c}} :G/N\rightarrow \mathbb {Z}_p(\theta _{/N})/p^n\mathbb {Z}_p(\theta _{/N})$ induces a continuous 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ satisfying $c\vert _N\equiv 0$ and $c={\bar{c}}\circ \pi$.

Proof of Proposition 2.10

Set ${\bar{G}}=G/N$ and ${\bar{\theta }}=\theta _{/N}$. For every $n\ge 1$, the canonical projection $\pi :G\rightarrow {\bar{G}}$ induces the inflation maps

$$\begin{aligned} \begin{aligned} f_n&:\textrm{H}^1({\bar{G}},\mathbb {Z}_p({\bar{\theta }} )/p^n\mathbb {Z}_p({\bar{\theta }} ))\longrightarrow \textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )), \\ f&:\textrm{H}^1({\bar{G}},\mathbb {Z}/p\mathbb {Z})\longrightarrow \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}), \end{aligned}\end{aligned}$$

(2.4)

which are injective by [20, Proposition 1.6.7]. Also, the epimorphisms (respectively, of continuous ${\bar{G}}$-modules and continuous G-modules) $\mathbb {Z}_p({\bar{\theta }} )/p^n\mathbb {Z}_p({\bar{\theta }} )\rightarrow \mathbb {Z}/p\mathbb {Z}$ and $\mathbb {Z}_p(\theta )/p^n\rightarrow \mathbb {Z}/p\mathbb {Z}$ induce, respectively, the morphisms

$$\begin{aligned} \begin{aligned} \tau _n^N&:\textrm{H}^1({\bar{G}},\mathbb {Z}_p(\theta )/p^n)\longrightarrow \textrm{H}^1({\bar{G}},\mathbb {Z}/p),\\ \tau _n&:\textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n)\longrightarrow \textrm{H}^1(G,\mathbb {Z}/p).\end{aligned}\end{aligned}$$

(2.5)

Altogether, by [20, Proposition 1.5.2] one has the commutative diagram

Since $(G,\theta )$ is Kummerian, $\tau _n$ is surjective for every $n\ge 1$. Given ${\bar{\beta }}\in \textrm{H}^1({\bar{G}},\mathbb {Z}/p\mathbb {Z})$, ${\bar{\beta }}\ne 0$, our goal is to find $\alpha \in \textrm{H}^1({\bar{G}},\mathbb {Z}_p({\bar{\theta }} )/p^n\mathbb {Z}_p({\bar{\theta }} ))$ such that ${\bar{\beta }}=\tau _n^N(\alpha )$.

Set $\beta ={\bar{\beta }}\circ \pi =f({\bar{\beta }})$. Then $\beta :G\rightarrow \mathbb {Z}/p$ is a non-trivial continuous homomorphism such that ${{\,\textrm{Ker}\,}}(\beta )\supseteq N$. By hypothesis, the morphism $N/N^p[G,N]\rightarrow G/\Phi (G)$ induced by the inclusion $N\hookrightarrow G$, and dual to $\textrm{res}_{G,N}^1$, is injective. Thus, one may find a minimal generating set $\mathcal {X}$ of G such that $\mathcal {Y}=\mathcal {X}\cap N$ generates N as a normal subgroup of G. By Lemma 2.9, there exists a continuous 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta )$ satisfying

$$\begin{aligned} c(x)\equiv \beta (x)\mod p\mathbb {Z}_p(\theta )\qquad \text {for every }x\in \mathcal {X}\end{aligned}$$

—i.e., $\tau _n([c])=\beta$, where $[c]\in \textrm{H}^1(G,\mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta ))$ denotes the cohomology class of c—and moreover $c(x)=0$ for every $x\in \mathcal {Y}$. Therefore, by Fact 2.11–(i), the restriction

$$\begin{aligned} c\vert _N:N\longrightarrow \mathbb {Z}_p(\theta )/p^n\mathbb {Z}_p(\theta ) \end{aligned}$$

is the map constantly equal to 0. By Fact 2.11–(ii), c induces a continuous 1-cocycle

$$\begin{aligned} {\bar{c}}:{\bar{G}} \longrightarrow \mathbb {Z}_p({\bar{\theta }})/p^n\mathbb {Z}_p({\bar{\theta }}) \end{aligned}$$

such that ${\bar{c}}\circ \pi =c$, and $[c]=f_n([{\bar{c}}])$, where $[{\bar{c}}]\in \textrm{H}^1({\bar{G}},\mathbb {Z}_p({\bar{\theta }})/p^n\mathbb {Z}_p({\bar{\theta }}))$ denotes the cohomology class of ${\bar{c}}$. Altogether, one has

$$\begin{aligned} f({\bar{\beta }})=\beta =\tau _n([c])=\tau _n\circ f_n([\bar{c}])=f\circ \tau _n^N([{\bar{c}}]).\end{aligned}$$

Since f is injective, one obtains ${\bar{\beta }}=\tau _n^N([{\bar{c}}])$. $\square$

Remark 2.12

Proposition 2.10 may be proved also in a purely group-theoretic way, see [3, Remark 3.9].

3 The $\mathbb {Z}/p\mathbb {Z}$-Cohomology of G

The purpose of this section is to prove the first statement of Proposition 1.3, and more in general to describe the $\mathbb {Z}/p\mathbb {Z}$-cohomology algebra $\textbf{H}^\bullet (G,\mathbb {Z}/p\mathbb {Z})$ with G as in Theorem 1.1.

3.1 Degree 1 and 2

Let G be a pro-p group. We set the subgroup $G_{(3)}$ of G as follows:

$$\begin{aligned} G_{(3)}={\left\{ \begin{array}{ll} G^p[G,G'] &{} \quad \text {if }p\ne 2,\\ G^4(G')^2[G,G'] &{} \quad \text {if }p=2, \end{array}\right. }\end{aligned}$$

i.e., $G_{(3)}$ is the third term of the p-Zassenhaus filtration of G (cf., e.g., [23, Sect. 3.1]). In particular, $G_{(3)}$ is a normal subgroup of the Frattini subgroup $\Phi (G)$, and the quotient $\Phi (G)/G_{(3)}$ is a p-elementary abelian pro-p group—and thus also a $\mathbb {Z}/p$-vector space.

Recall that the cohomology group $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ is equal to the group of pro-p group homomorphisms from G to $\mathbb {Z}/p$, namely, one has

$$\begin{aligned} \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})=\textrm{Hom}(G,\mathbb {Z}/p\mathbb {Z})\simeq (G/\Phi (G))^*, \end{aligned}$$

(3.1)

where $\textvisiblespace ^*$ denotes the $\mathbb {Z}/p$-dual (cf., e.g., [32, Chap. I, Sect. 4.2]). Thus, the dimension of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ is equal to the cardinality $\textrm{d}(G)$ of any minimal generating set of G. On the other hand, the dimension of $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$ is equal to the number $\textrm{r}(G)$ of defining relations of G (cf. [32, Chap. I, Sect. 4.3]). Moreover, if both $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ and $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$ are finite, and if the cup-product yields an epimorphism $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})^{\otimes 2}\twoheadrightarrow \textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$, one has an isomorphism of elementary abelian p-groups

(3.2)

(cf. [17, Theorem 7.3]). For further properties of the cohomology of pro-p groups we refer to [32, Chap. I, Sect. 4] and to [20, Chap. III, Sect. 9].

3.2 Amalgams

Henceforth, G will denote a pro-p group as in Theorem 1.1. Set

$$\begin{aligned} \begin{aligned} G_1&= \langle \,x,y_1,\ldots ,y_{d_1}\,\mid \,x^{\epsilon p}[x,y_1]\cdots [y_{d_1-1},y_{d_1}]=1\,\rangle ,\\ G_2&= \langle \,x,z_1,\ldots ,z_{d_2}\,\mid \,x^{\epsilon p}[x,z_1]\cdots [z_{d_2-1},z_{d_2}]=1\,\rangle , \end{aligned} \end{aligned}$$

with $\epsilon =0,1$ depending on whether we are considering case (1.1.a) or (1.1.b). Then $G_1,G_2$ are Demushkin groups, and G is the amalgamated free pro-p product

$$\begin{aligned} G=G_1\amalg _X^{{\hat{p}}}G_2, \end{aligned}$$

(3.3)

with amalgam the subgroup $X\subseteq G_1,G_2$ generated by x. Observe that $X\simeq \mathbb {Z}_p$, as X has infinite index in both $G_1,G_2$, and subgroups of infinite index of Demushkin groups are free pro-p groups (cf. [32, Chap. I, Sect. 4.5, Example 5–(b)]). Therefore, the amalgamated free pro-p product is proper, i.e., $G_1,G_2\subseteq G$ (cf. [31]).

3.3 Quadratic Cohomology

Let

$$\begin{aligned} \mathcal {B}=\left\{ \,\chi ,\,\varphi _1,\,\ldots ,\,\varphi _{d_1},\,\psi _1,\,\ldots ,\,\psi _{d_2}\,\right\} \end{aligned}$$

be the basis of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})=\textrm{Hom}(G,\mathbb {Z}/p\mathbb {Z})$ dual to $\mathcal {X}=\{x,y_1,\ldots ,z_{d_2}\}$—i.e.,

$$\begin{aligned}\begin{aligned}&\chi (w)={\left\{ \begin{array}{ll} 1&{}\text {\quad if }w=x \\ 0 &{}\quad \text {if }w=y_i,z_j \end{array}\right. }\qquad \text {and}\\&\varphi _i(w)={\left\{ \begin{array}{ll} \delta _{i,i'}&{}\quad \text {if }w=y_{i'} \\ 0 &{}\quad \text {if }w=x,z_j, \end{array}\right. }\qquad \psi _j(w)={\left\{ \begin{array}{ll} \delta _{j,j'}&{}\quad \text {if }w=z_{j'} \\ 0 &{}\quad \text {if }w=x,y_i, \end{array}\right. } \end{aligned}\end{aligned}$$

for every $1\le i,i'\le d_1$ and $1\le j,j'\le d_2$ (cf. (3.1)). With an abuse of notation, we may consider the subsets $\mathcal {B}_1=\{\chi ,\varphi _1,\ldots ,\varphi _{d_1}\}$, $\mathcal {B}_2=\{\chi ,\psi _1,\ldots ,\psi _{d_2}\}$, and $\mathcal {B}_X=\{\chi \}$, as bases of $\textrm{H}^1(G_1,\mathbb {Z}/p\mathbb {Z})$, $\textrm{H}^1(G_2,\mathbb {Z}/p\mathbb {Z})$, and $\textrm{H}^1(X,\mathbb {Z}/p\mathbb {Z})$ respectively.

Proposition 3.1

The algebra $\textbf{H}^\bullet (G,\mathbb {Z}/p\mathbb {Z})$ is quadratic.

Proof

As stated in Sect. 3.2, $G=G_1\amalg _{X}^{{\hat{p}}}G_2$ is a proper amalgamated free pro-p product. Since $\mathcal {B}_X\subseteq \mathcal {B}_1,\mathcal {B}_2$, the restriction maps

$$\begin{aligned} {{\,\textrm{res}\,}}^1_{G_i,X}:\textrm{H}^1(G_i,\mathbb {Z}/p\mathbb {Z})\longrightarrow \textrm{H}^1(X,\mathbb {Z}/p\mathbb {Z}),\qquad \text {with }i=1,2, \end{aligned}$$

are surjective. Moreover, $\textrm{H}^2(X,\mathbb {Z}/p\mathbb {Z})=0$, as $X\simeq \mathbb {Z}_p$, and thus ${{\,\textrm{Ker}\,}}({{\,\textrm{res}\,}}_{G_i,X}^2)=\textrm{H}^2(G_i,\mathbb {Z}/p\mathbb {Z})$ for both $i=1,2$. On the other hand, $\textrm{H}^1(G_1,\mathbb {Z}/p\mathbb {Z})$ and $\textrm{H}^1(G_2,\mathbb {Z}/p\mathbb {Z})$ are generated by $\chi \smallsmile \varphi _1$ and $\chi \smallsmile \psi _1$ respectively, as $G_1,G_2$ are Demushkin groups (cf., e.g., [20, Proposition 3.9.16]), and thus

$$\begin{aligned}{} & {} {{\,\textrm{Ker}\,}}({{\,\textrm{res}\,}}_{G_i,X}^2) \\{} & {} =\textrm{H}^2(G_i,\mathbb {Z}/p\mathbb {Z})={{\,\textrm{Ker}\,}}({{\,\textrm{res}\,}}_{G_i,X}^1)\smallsmile \textrm{H}^1(G_i,\mathbb {Z}/p\mathbb {Z}),\qquad \text {with }i=1,2, \end{aligned}$$

as ${{\,\textrm{res}\,}}_{G_1,X}^1(\varphi _1)=0$ and ${{\,\textrm{res}\,}}_{G_2,X}^1(\psi _1)=0$. Finally, Demushkin groups are well-known to yield a quadratic $\mathbb {Z}/p\mathbb {Z}$-cohomology algebra, while $\textbf{H}^\bullet (X,\mathbb {Z}/p\mathbb {Z})$ is obviously quadratic, as $X\simeq \mathbb {Z}_p$. Therefore, we may apply [28, Theorem B], so that also $\textbf{H}^\bullet (G,\mathbb {Z}/p\mathbb {Z})$ is quadratic. $\square$

We describe now more in detail the structure of $\textbf{H}^\bullet (X,\mathbb {Z}/p\mathbb {Z})$. By duality—cf. [17, Theorem 7.3] and (3.2)—the set $\{\chi \smallsmile \varphi _1,\chi \smallsmile \psi _1\}$ is a basis of $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$, and in $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$ one has the relations

$$\begin{aligned} \chi \smallsmile \varphi _{i'}=\chi \smallsmile \psi _{j'}=\varphi _i\smallsmile \psi _j=0 \end{aligned}$$

(3.4)

for all $1\le i,i'\le d_1$ and $1\le j,j'\le d_2$, with $i',j'\ne 1$, and

$$\begin{aligned} \begin{aligned} \varphi _{i}\smallsmile \varphi _{i'}={\left\{ \begin{array}{ll} (-1)^\epsilon \chi \smallsmile \varphi _1 &{} \quad \text {if }2\mid i=i'-1, \\ 0&{}\quad \text {otherwise}, \end{array}\right. }\\ \qquad \psi _{j}\smallsmile \psi _{j'}={\left\{ \begin{array}{ll} (-1)^\epsilon \chi \smallsmile \psi _1 &{} \quad \text {if }2\mid j=j'-1, \\ 0&{}\quad \text {otherwise} \end{array}\right. } \end{aligned}\end{aligned}$$

(3.5)

(see also [23, Sect. 3.2]).

Finally, one has an exact sequence

(cf. [28, p. 653]). Since $\textrm{H}^2(X,\mathbb {Z}/p\mathbb {Z})=\textrm{H}^3(G_i,\mathbb {Z}/p\mathbb {Z})=0$ for both $i=1,2$, one has $\textrm{H}^3(G,\mathbb {Z}/p\mathbb {Z})=0$, and thus by quadraticity also $\textrm{H}^n(G,\mathbb {Z}/p\mathbb {Z})=0$ for all $n\ge 3$.

Remark 3.2

It is well-known that if a pro-p group has non-trivial torsion, then its nth $\mathbb {Z}/p$-cohomology group is non-trivial for every $n>0$; hence, G is torsion-free.

4 Proof of Theorem 1.1 Case (1.1.a)

Let G be a pro-p group as defined in Theorem 1.1, with defining relations as in (1.1.a)—namely,

$$\begin{aligned} G=\langle \,x,y_1,\ldots ,y_{d_1},z_1,\ldots ,z_{d_2}\,\mid \,r_1=r_2=1\,\rangle ,\end{aligned}$$

with $d_1+d_2\ge 4$ and

$$\begin{aligned}\begin{aligned} r_1&= [x,y_1]\cdots [y_{d_1-1},y_{d_1}],\\r_2&=[x,z_1]\cdots [z_{d_2-1},z_{d_2}]. \end{aligned}\end{aligned}$$

Without loss of generality, we may assume that $d_1\ge 3$.

4.1 Kummerianity

Let $G_1,G_2$ be the two Demushkin groups as in Sect. 3.2, with $\epsilon =0$. By Example 2.6, if

$$\begin{aligned} \theta _1:G_1\longrightarrow 1+p\mathbb {Z}_p\qquad \text {and} \qquad \theta _2:G_2\longrightarrow 1+p\mathbb {Z}_p \end{aligned}$$

are two torsion-free orientations completing respectively $G_1$ and $G_2$ into Kummerian oriented pro-p groups, then necessarily $\theta _1(x)=\theta _1(y_1)=\cdots =\theta _1(y_{d_1})=1$, and analogously $\theta _2(x)=\theta _2(z_1)=\cdots =\theta _1(z_{d_2})=1$.

Proposition 4.1

Let $\theta :G\rightarrow 1+p\mathbb {Z}_p$ be a torsion-free orientation. Then the oriented pro-p group $(G,\theta )$ is Kummerian if, and only if, $\theta$ is constantly equal to 1.

Proof

If $\theta \equiv \textbf{1}$, then $(G,\textbf{1})$ is Kummerian if, and only if, the abelianization $G^{{{\,\textrm{ab}\,}}}$ is a free abelian pro-p group. But this is easily verified, as clearly $G^{{{\,\textrm{ab}\,}}}\simeq \mathbb {Z}_p^{d_1+d_2-1}$.

Conversely, suppose that $(G,\theta )$ is Kummerian. Let $N_1$ and $N_2$ denote the normal subgroups of G generated as normal subgroups by $z_1,\ldots ,z_{d_2}$ and $y_1,\ldots ,y_{d_1}$ respectively. Then $G/N_1\simeq G_1$ and $G/N_2\simeq G_2$. Moreover, Proposition 2.10 implies that $(G/N_i,\theta _{/N_i})$ is Kummerian for both $i=1,2$. Since $G/N_i\simeq G_i$ for both i, Example 2.6 and the argument before the statement of the proposition imply that the torsion-free orientations $\theta _{/N_1}$ and $\theta _{/N_2}$ are constantly equal to 1. Hence, also $\theta$ is constantly equal to 1, as $\theta (w)=\theta _{/N_1}(wN_1)$ for every $w\in G_1$, and analogously $\theta (w)=\theta _{/N_2}(wN_2)$ for every $w\in G_2$. $\square$

Therefore, if G may complete into a 1-cyclotomic oriented pro-p group, then necessarily G is absolutely torsion-free. In order to prove Theorem 1.1 in case (1.1.a), we aim at exhibiting an open subgroup H of G, of index $p^2$, whose abelianization $H^{{{\,\textrm{ab}\,}}}$ has non-trivial torsion.

4.2 The Subgroup U

Set $u=y_3^p$, $t_0=z_1^{-1}y_3$, and $t_h=t_0t_0^{y_3}\cdots t_0^{y_3^h}$ for all $h=0,\ldots ,p-1$. A straightforward computation shows that

$$\begin{aligned} z_1^{h}=y_3^h\cdot (t_0^{-1})^{y_3^{h-1}}\cdots (t_0^{-1})^{y_3}\cdot t_0^{-1}=y_3^ht_{h-1}^{-1} \end{aligned}$$

(4.1)

for all $h=0,\ldots ,p-1$.

Let $\phi _G:G\rightarrow \mathbb {Z}/p$ be the homomorphism of pro-p groups defined by $\phi _G(y_3)=\phi _G(z_1)=1$ and $\phi _G(x)=\phi _G(y_i)=\phi _G(z_j)=0$ for all $i=1,2,4,\ldots ,d_1$ and $j=2,\ldots ,d_2$, and set $U={{\,\textrm{Ker}\,}}(\phi )$. Then U is an open subgroup of G of index p, generated as a normal subgroup by the subset

$$\begin{aligned} \mathcal {X}=\left\{ \, u,\, x,\, t_0,\, y_i,\,z_j\,\mid \, i=1,2,4,\ldots ,d_1,\,j=2,\ldots ,d_{2} \right\} , \end{aligned}$$

and $G/U=\{U,y_3U,\ldots ,y_3^{p-1}U\}$.

Lemma 4.2

The subset

$$\begin{aligned} \mathcal {Y}_U=\left\{ \,u,\,x,\,y_2,\,t_h,\,y_i^{y_3^h},\,z_j^{y_3^h}\,\mid \,i=1,4,\ldots ,d_1,\,j=2,\ldots ,d_{2},\,h=0,\ldots ,p-1\,\right\} \end{aligned}$$

of U is a minimal generating set of U as a pro-p group.

Proof

Since U is normally generated by $\mathcal {X}$ and $G/U=\{U,\ldots ,y_3^{p-1}U\}$, U is generated as a pro-p group by the set $\{w^{y_3^h}\mid w\in \mathcal {X},h=0,\ldots ,p-1 \}$. Also, U is subject to the relations

$$\begin{aligned} r_1^{y_3^h}= & {} \left[ x^{y_3^h},y_1^{y_3^h}\right] \cdots \left[ y_{d_1-1}^{y_3^h},y_{d_1}^{y_3^h}\right] =1, \end{aligned}$$

(4.2)

$$\begin{aligned} r_2^{y_3^h}= & {} \left[ x^{y_3^h},z_1^{y_3^h}\right] \cdots \left[ z_{d_2-1}^{y_3^h},z_{d_2}^{y_3^h}\right] =1, \end{aligned}$$

(4.3)

with $h=0,\ldots ,p-1$.

Consider the abelianization $U^{{{\,\textrm{ab}\,}}}$. Since the only factor in (4.2) which does not lie in $U'$ is $[y_2^{y_3^h},y_3]$, the relation (4.2) implies that $[y_2^{y_3^h},y_3]\in U'$ as well, and therefore

$$\begin{aligned} y_2^{y_3^h}\equiv y_2\mod U'\qquad \text {for all }h=0,\ldots ,p-1. \end{aligned}$$

Analogously, the only factor in (4.3) which does not lie in $U'$ is $[x^{y_3^h},z_1^{y_3^h}]$, so that the relation (4.2) implies that $[x^{y_3^h},z_1^{y_3^h}]\in U'$ as well. Hence, one has

$$\begin{aligned}\begin{aligned}{}[x,z_1]\equiv 1\bmod U'\;&\Rightarrow \; x^{y_3t_0^{-1}}\equiv x\bmod U'\\&\Rightarrow x^{y_3}\equiv x^{t_0}\bmod U',\\ \left[ x^{y_3},z_1^{y_3}\right] \equiv 1\bmod U' \;&\Rightarrow \; (x^{y_3})^{(z_1^{y_3})}=x^{y_3^2(t_0^{-1})^{y_3}}\equiv x^{y_3}\bmod U'\\&\Rightarrow \; x^{y_3^2}\equiv x^{t_1}\bmod U', \end{aligned}\end{aligned}$$

and so on. Thus,

$$\begin{aligned} x^{y_3^h}\equiv x^{t_{h-1}}\mod U'\qquad \text {for all }h=1,\ldots ,p-1. \end{aligned}$$

Altogether, $U^{{{\,\textrm{ab}\,}}}$ is the free abelian pro-p group generated by the cosets $\{wU'\mid w\in \mathcal {Y}_U\}$, so that Fact 2.1 yields the claim. $\square$

Now set $U_1=G_1\cap U$ and $U_2=G_2\cap U$. Then $U_1,U_2$ are open subgroups of $G_1,G_2$, respectively, of index p, and thus they are again Demushkin groups, on $2+p(d_1-1)$ and $2+p(d_2-1)$ generators respectively (cf. [6]). In particular, the defining relation of $U_1$ is

$$\begin{aligned} s_1= \prod _{h=p-1}^0\left( \left[ y_{4}^{y_3^h},y_{5}^{y_3^h}\right] \cdots \left[ y_{d_1-1}^{y_3^h},y_{d_1}^{y_3^h}\right] \left[ x^{y_3^h},y_{1}^{y_3^h}\right] \right) [y_2,u]=1, \end{aligned}$$

(4.4)

while the defining relation of $U_2$ is

$$\begin{aligned} \begin{aligned} s_2&=\prod _{h=p-1}^0\left( \left[ z_{2}^{z_1^h},z_{3}^{z_1^h}\right] \cdots \left[ z_{d_2-1}^{z_1^h},z_{d_2}^{z_1^h}\right] \right) [x,z_1^p] \\&=\prod _{h=p-1}^0\left( \left[ z_{2}^{y_3^ht_{h-1}^{-1}},z_{3}^{y_3^ht_{h-1}^{-1}}\right] \cdots \left[ z_{d_2-1}^{y_3^ht_{h-1}^{-1}},z_{d_2}^{y_3^ht_{h-1}^{-1}}\right] \right) [x,ut_{p-1}^{-1}] =1. \end{aligned} \end{aligned}$$

(4.5)

Also, from the relations (4.4)–(4.5) and from (4.1), one computes

$$\begin{aligned} \begin{aligned} x^{y_3}&=x^{z_1t_0}=x^{t_0}([z_{d_2},z_{d_2-1}]\cdots [z_3,z_2])^{t_0},\\ x^{y_3^2}&= x^{t_1}([z_{d_2},z_{d_2-1}]\cdots )^{t_1} \left( \left[ z_{d_2}^{y_3},z_{d_2-1}^{y_3}\right] \cdots \right) ^{t_0^{-1}t_1},\\ x^{y_3^3}&= x^{t_2}([z_{d_2},z_{d_2-1}]\cdots )^{t_2} \left( \left[ z_{d_2}^{y_3},z_{d_2-1}^{y_3}\right] \cdots \right) ^{t_0^{-1}t_2} \left( \left[ z_{d_2}^{y_3^2},z_{d_2-1}^{y_3^2}\right] \cdots \right) ^{t_1^{-1}t_2}, \end{aligned} \end{aligned}$$

(4.6)

and so on. In fact, the two relations (4.4)–(4.5)—with the $x^{y_3^h}$’s replaced using (4.6)—are all the defining relations we need to get U, as shown in the following.

Lemma 4.3

The pro-p group U has $\textrm{r}(U)=2$ defining relations.

Proof

Since $\textrm{H}^n(G,\mathbb {Z}/p\mathbb {Z})=0$ for every $n\ge 3$ (cf. Sect. 3.3) and $[G:U]=p$, one has $\textrm{H}^n(U,\mathbb {Z}/p\mathbb {Z})=0$ for every $n\ge 3$ as well (cf. [20, Proposition 3.3.5]). Moreover, one has

$$\begin{aligned} \textrm{r}(U)-\textrm{d}(U)+1=p\left( \textrm{r}(G)-\textrm{d}(G)+1\right) \end{aligned}$$

(4.7)

(cf. [20, Proposition 3.3.13]). By definition, $\textrm{r}(G)=2$ and $\textrm{d}(G)=1+d_1+d_2$, while $\textrm{d}(U)=3+p(d_1+d_2-2)$ by Lemma 4.2. Therefore, from (4.7) one computes $\textrm{r}(U)=2$. $\square$

4.3 The Subgroup H

Let $\phi _U:U\rightarrow \mathbb {Z}/p$ be the homomorphism of pro-p groups defined by $\phi _U(y_1)$, $\phi _U(y_1^{y_3})=-1$, and $\phi _U(w)=0$ for any other element w of $\mathcal {Y}_U$, and put $H={{\,\textrm{Ker}\,}}(\phi _U)$. Then H is an open subgroup of U of index p. Set $v=y_1$. Since $U/H=\{H,vH,\ldots ,v^{p-1}H\}$, H is the pro-p group (non-minimally) generated by

$$\begin{aligned} \mathcal {X}_{H}=\left\{ \,v^p,\,\left( vy_1^{y_3}\right) ^{v^h},\,w^{v^h}\,\mid w\in \mathcal {Y}_U,\,w\ne v,y_1^{y_3},\, h=0,\ldots ,p-1\,\,\right\} , \end{aligned}$$

and subject to the 2p relations $s_1^{v^h}=1$ and $s_2^{v^h}=1$, with $h=0,\ldots ,p-1$. We claim that the abelianization $H^{{{\,\textrm{ab}\,}}}$ yields non-trivial torsion.

Proposition 4.4

The abelian pro-p group $H^{{{\,\textrm{ab}\,}}}$ is not torsion-free.

Proof

Since all the elements of $\mathcal {Y}_U$ showing up in the last terms of the equalities (4.6) belong to H, one deduces that $x^{y_3^h}\equiv x\bmod H'$ for all $h=0,\ldots ,p-1$.

Now, each factor of $s_2$—cf. (4.5)—is a commutator of elements of H, and thus the relations $s_2^{v^h}=1$ yield trivial relations in $H^{{{\,\textrm{ab}\,}}}$. On the other hand, every factor of $s_1$—cf. (4.4)—but $[x,y_1]$ and $[x^{y_3},y_1^{y_3}]$, is a commutator of elements of H. From (4.4) one obtains

$$\begin{aligned} \left[ x^{y_3},y_1^{y_3}\right] [x,y_1]\equiv \left[ x,v^{-1}(vy_1^{y_3})\right] [x,v]\equiv [x,v^{-1}][x,v]\equiv 1\mod H', \end{aligned}$$

(4.8)

as $vy_1^{y_3}\in H$. Altogether, $H^{{{\,\textrm{ab}\,}}}$ is the abelian pro-p group (non-minimally) generated by the set $\mathcal {X}_{H^{{{\,\textrm{ab}\,}}}}=\{wH'\,\mid \,w\in \mathcal {X}_H\}$, and subject to the p relations

$$\begin{aligned} \left[ x^{v^h}H',v^{-1}H'\right] \left[ x^{v^h}H',vH'\right] =H',\qquad \text {with }h=0,\ldots ,p-1, \end{aligned}$$

as $U/H=\{H,vH,\ldots ,v^{p-1}H\}$. From these relations, one deduces the equivalences:

$$\begin{aligned} \begin{aligned} x^{v^2}&\equiv \left( x^v\right) ^2\cdot x^{-1}\mod H'\qquad \text {with }h=1,\\ x^{v^3}&\equiv \left( x^{v^2}\right) ^2\cdot \left( x^v\right) ^{-1}\equiv \left( x^v\right) ^3\cdot x^{-2} \mod H'\qquad \text {with }h=2,\\&\quad \vdots \\ x^{v^{p-1}}&\equiv \left( x^{v^{p-2}}\right) ^2\cdot \left( x^{v^{p-3}}\right) ^{-1}\equiv \left( x^v\right) ^{p-1}\cdot x^{2-p} \mod H'\qquad \text {with }h=p-2,\\ x^{v^{p}}&\equiv \left( x^{v^{p-1}}\right) ^2\cdot \left( x^{v^{p-2}}\right) ^{-1}\equiv \left( x^v\right) ^{p}\cdot x^{1-p} \mod H'\qquad \text {with }h=p-1.\\ \end{aligned} \end{aligned}$$

But $x^{v^p}\equiv x\bmod H'$, as $v^p\in H$, and thus from the last of the above equivalences one obtains

$$\begin{aligned} x\equiv (x^v)^px^{1-p}\mod H'\;\Longrightarrow \;(x^v)^px^{-p}\equiv (x^vx^{-1})^p\equiv 1\mod H'. \end{aligned}$$

(4.9)

Altogether, $H^{{{\,\textrm{ab}\,}}}$ is the abelian pro-p group minimally generated by

$$\begin{aligned} \mathcal {Y}_{H^{{{\,\textrm{ab}\,}}}}=\left\{ \,v^pH',\, xH', \, x^vH',\,\left( vy_1^{y_3}\right) ^{v^h}H',\,w^{v^h}H'\,\mid \, h=0,\ldots ,p-1\,\,\right\} , \end{aligned}$$

where $w\in \mathcal {Y}_U\smallsetminus \{v,y_1^{y_3},x\}$, and subject to the relation $((xH')^{-1}\cdot x^vH')^p=H'$—in particular, $H^{{{\,\textrm{ab}\,}}}$ is isomorphic to $\mathbb {Z}_p^{2+p+p^2(d_1+d_2-2)}\times \mathbb {Z}/p\mathbb {Z}$. $\square$

5 Proof of Theorem 1.1 Case (1.1.b)

Let p be an odd prime, and let G be a pro-p group as defined in Theorem 1.1, with defining relations as in (1.1.b)—namely,

$$\begin{aligned} G=\langle \,x,y_1,\ldots ,y_{d_1},z_1,\ldots ,z_{d_2}\,\mid \,r_1=r_2=1\,\rangle ,\end{aligned}$$

with

$$\begin{aligned}\begin{aligned} r_1&= y_1^p[y_1,x]\cdots [y_{d_1-1},y_{d_1}],\\r_2&=z_1^p[z_1,x]\cdots [z_{d_2-1},z_{d_2}]. \end{aligned}\end{aligned}$$

5.1 Kummerianity

Let $G_1,G_2$ be the two Demushkin groups as in Sect. 3.2, with $\epsilon =1$. By Example 2.6, if

$$\begin{aligned} \theta _1:G_1\longrightarrow 1+p\mathbb {Z}_p\qquad \text {and} \qquad \theta _2:G_2\longrightarrow 1+p\mathbb {Z}_p \end{aligned}$$

are two torsion-free orientations completing respectively $G_1$ and $G_2$ into Kummerian oriented pro-p groups, then necessarily $\theta _1(y_1)=\cdots =\theta _1(y_{d_1})=1$, and analogously $\theta _2(z_1)=\cdots =\theta _1(z_{d_2})=1$, while $\theta _1(x)=\theta _2(x)=(1-p)^{-1}$.

Proposition 5.1

An orientation $\theta :G\rightarrow 1+p\mathbb {Z}_p$ completes G into a Kummerian oriented pro-p group $(G,\theta )$ if, and only if,

$$\begin{aligned} \theta (x)=(1-p)^{-1}\qquad \text {and}\qquad \theta (y_i)=\theta (z_j)=1 \end{aligned}$$

for all $i=1,\ldots ,d_1$ and $j=1,\ldots ,d_2$.

Proof

Suppose that $\theta :G\rightarrow 1+p\mathbb {Z}_p$ is the orientation defined as above, and pick arbitrary p-adic integers $\lambda ,\lambda _i,\lambda '_j\in \mathbb {Z}_p$ for $1\le i\le d_1$ and $1\le j\le d_2$. The assignment $x\mapsto \lambda$, $y_i\mapsto \lambda _i$ and $z_j\mapsto \lambda '_j$ for every i, j yields a well-defined continuous 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )$, as (2.3) implies that

$$\begin{aligned}\begin{aligned} c(r_1)&= c(y_1^p)+c([y_1,x])+c([y_2,y_3])+\cdots +c([y_{d_1-1},y_{d_1}])\\&= p\cdot \lambda _1+ \theta (x)^{-1}(\lambda _1(1-\theta (x))-0)+0+\cdots +0 \\ {}&=0 \end{aligned}\end{aligned}$$

and

$$\begin{aligned}\begin{aligned} c(r_2)&= c(z_1^p)+c([z_1,x])+c([z_2,z_3])+\cdots +c([z_{d_2-1},z_{d_2}])\\&= p\cdot \lambda '_1+ \theta (x)^{-1}(\lambda '_1(1-\theta (x))-0)+0+\cdots +0 \\ {}&=0. \end{aligned}\end{aligned}$$

Therefore, $(G,\theta )$ is Kummerian by Lemma 2.9.

Conversely, suppose that $(G,\theta )$ is Kummerian. Let $N_1$ and $N_2$ denote the normal subgroups of G generated as normal subgroups by $z_1,\ldots ,z_{d_2}$ and $y_1,\ldots ,y_{d_1}$, respectively. Then $G/N_1\simeq G_1$ and $G/N_2\simeq G_2$. Moreover, Proposition 2.10 implies that $(G/N_i,\theta _{/N_i})$ is Kummerian for both $i=1,2$.

Since $G/N_i\simeq G_i$ for both i, Example 2.6 and the argument before the statement of the proposition imply that $\theta _{/N_1}(y_1N_1)=\cdots =\theta _{/N_1}(y_{d_1}N_1)=1$, and analogously $\theta _{/N_2}(z_1N_2)=\cdots =\theta _{/N_2}(z_{d_2}N_2)=1$, while $\theta _{/N_1}(xN_1)=\theta _{/N_2}(xN_2)=(1-p)^{-1}$. Hence, $\theta$ is as defined above, as $\theta (w)=\theta _{/N_1}(wN_1)$ for every $w\in G_1$, and analogously $\theta (w)=\theta _{/N_2}(wN_2)$ for every $w\in G_2$. $\square$

Henceforth, $\theta :G\rightarrow 1+p\mathbb {Z}_p$ will denote the orientation as in Proposition 5.1.

5.2 The Subgroup H

Let $\phi _1:G_1\rightarrow \mathbb {Z}/p\oplus \mathbb {Z}/p$ and $\phi _2:G_2\rightarrow \mathbb {Z}/p\oplus \mathbb {Z}/p$ be the homomorphisms of pro-p groups defined by

$$\begin{aligned} \begin{aligned}&\phi _1(x)=\phi _2(x)=(1,0),\\ {}&\phi _1(y_1)=\phi _2(z_1)=(0,1),\\ {}&\phi _1(y_i)=\phi _2(z_j)=(0,0)\text { for }i,j\ge 2. \end{aligned} \end{aligned}$$

(5.1)

Put $U_1={{\,\textrm{Ker}\,}}(\phi _1)$ and $U_2={{\,\textrm{Ker}\,}}(\phi _2)$, and also

$$\begin{aligned} t=z_1^{-1}y_1,\qquad u=x^p,\qquad v=y_1^p,\qquad w=z_1^p. \end{aligned}$$

Then $U_1$ is an open normal subgroup of $G_1$ of index $p^2$, and likewise for $U_2$ and $G_2$—note that by [6] both $U_1$ and $U_2$ are Demushkin groups.

Finally, put $N_1={{\,\textrm{Ker}\,}}(\theta \vert _{U_1})$, $N_2={{\,\textrm{Ker}\,}}(\theta \vert _{U_2})$, and let T be the subgroup of G generated by t. Observe that $N_1$ and $N_2$ are free pro-p groups, as they are subgroups of infinite index of Demushkin groups (cf. [32, Chap. I, Sect. 4.5, Example 5–(b)]), while $T\simeq \mathbb {Z}_p$ as G is torsion-free (cf. Remark 3.2).

Let H be the subgroup of G generated by $U_1$, $U_2$ and T, and let M be the subgroup of H generated by $N_1$, $N_2$ and T. Observe that $M\subseteq {{\,\textrm{Ker}\,}}(\theta )$. Our aim is to show that the oriented pro-p group $(H,\theta \vert _H)$ is not Kummerian. For this purpose, we need the following.

Lemma 5.2

• (i) $M=N_1\amalg N_2\amalg T$.

• (ii) M is a normal subgroup of H, and $H\simeq M\rtimes X^p$

• (iii) One has an isomorphism of p-elementary abelian groups

  $$\begin{aligned} \frac{G}{\Phi (G)}\simeq \frac{X^p}{X^{p^2}}\times \frac{N_1}{N_1^p[N_1,U_1]}\times \frac{N_2}{N_2^p[N_2,U_2]}\times \frac{T}{T^p}. \end{aligned}$$

  (5.2)

Proof

Consider the pro-p tree $\mathcal {T}$ associated to the amalgamated free pro-p product (3.3). Namely, $\mathcal {T}$ consists of a set vertices $\mathcal {V}$ and a set of edges $\mathcal {E}$, where

$$\begin{aligned}\begin{aligned} \mathcal {V}&=\{\,hG_1,hG_2\,\mid \,h\in G\,\}=G/G_1\,\dot{\cup }\, G/G_2,\\ \mathcal {E}&=\{\,hX\,\mid \,h\in G\,\}=G/X, \end{aligned}\end{aligned}$$

and it comes endowed with a natural G-action, i.e.,

$$\begin{aligned} \begin{aligned} g.(hG_1)=(gh)G_1\qquad&\text {for every } g\in G,\, hG_1\in G/G_1\subseteq \mathcal {V}\\ g.(hG_1)=(gh)G_2 \qquad&\text {for every }g\in G,\,hG_2\in G/G_2\subseteq \mathcal {V},\\ g.(hX)=(gh)X\qquad&\text {for every }g\in G,\,hX\in G/X=\mathcal {E}. \end{aligned}\end{aligned}$$

(5.3)

Pick $g\in M$ and $hX\in \mathcal {E}$. Then $g.hX=hX$ if, and only if, $g\in hXh^{-1}$, i.e., $g=hx^\lambda h^{-1}$ for some $\lambda \in \mathbb {Z}_p$. Since $M\subseteq {{\,\textrm{Ker}\,}}(\theta )$, it follows that

$$\begin{aligned} 1=\theta (g)=\theta \left( hx^\lambda h^{-1}\right) =\theta (x)^\lambda =(1-p)^{\lambda }, \end{aligned}$$

(5.4)

and therefore $\lambda =0$, as $1+p\mathbb {Z}_p$ is torsion-free. Hence, the subgroup M intersects trivially the stabilizer $\textrm{Stab}_{G}(hX)$ of every edge $hX\in \mathcal {E}$. By [14, Theorem 5.6], M decomposes as free pro-p product as follows:

$$\begin{aligned} M=\left( \coprod _{\omega \in \mathcal {V}'}\textrm{Stab}_{M}(\omega )\right) \amalg F, \end{aligned}$$

(5.5)

where F is a free pro-p group, and $\mathcal {V}'\subseteq \mathcal {V}$ is a continuous set of representatives of the space of orbits $M\backslash \mathcal {V}$. Clearly, the vertices $G_1$ and $G_2$ belong to different orbits, thus in the decomposition (5.5) one finds the two factors

$$\begin{aligned} \begin{aligned} \textrm{Stab}_M(G_1)=\{\,g\in M\,\mid \,gG_1=G_1\,\}=M\cap G_1, \\ \textrm{Stab}_M(G_2)=\{\,g\in M\,\mid \,gG_2=G_2\,\}=M\cap G_2. \end{aligned}\end{aligned}$$

Since $N_1\subseteq M\cap G_1\subseteq {{\,\textrm{Ker}\,}}(\theta )\cap G_1=N_1$, one has $\textrm{Stab}_M(G_1)=N_1$, and analogously $\textrm{Stab}_M(G_2)=N_2$. Therefore, from (5.5) one obtains

$$\begin{aligned} M=N_1\amalg N_2\amalg \left( \coprod _{\omega \in \mathcal {V}'\smallsetminus \{G_1,G_2\}}\textrm{Stab}_{M}(\omega )\amalg F\right) . \end{aligned}$$

(5.6)

It is straightforward to see that $t\notin N_1\amalg N_2$. Since M is generated as pro-p group by $N_1$, $N_2$ and t, the right-side factor in (5.6) is necessarily T, and this proves (i).

To prove (ii), we need only to show that $uMu^{-1}=M$, as $H=\langle \,u,M\,\rangle$. Since $N_1$ is normal in $U_1$, and $u\in U_1$, then $uN_1u^{-1}=N_1$—analogously, $uN_2u^{-1}=N_2$. Now, observe that the integer

$$\begin{aligned} (1-p)^p-1=\left( 1-\left( {\begin{array}{c}p\\ 1\end{array}}\right) p+\left( {\begin{array}{c}p\\ 2\end{array}}\right) p^2-\cdots -p^p\right) -1 \end{aligned}$$

is divisible by $p^2$ but not by $p^3$, so we put $(1-p)^p=1+p^2\lambda$, with $\lambda \in 1+p\mathbb {Z}_p$. From the relation $r_1=1$ one deduces

$$\begin{aligned} y_1^{x}=y_1^{1-p}\cdot \left( [y_2,y_3]\cdots [y_{d_1-1},y_{d_1}]\right) ^{-1}, \end{aligned}$$

(5.7)

and by iterating (5.7) p times, one obtains $y_1^{u}=y_1^{(1-p)^p} n_1$ for some $n_1\in N_1'$—for this purpose, observe that for every $\nu \ge 0$ and $i\ge 1$, the triple commutator

$$\begin{aligned} \left[ y_1^\nu ,[y_i,y_{i+1}]\right] =\left[ y_i^{y_1^\nu },y_{i+1}^{y_1^\nu }\right] ^{-1}\cdot [y_i,y_{i+1}] \end{aligned}$$

belongs to $N_1'$, as $y_i^{y_0^\nu }\in N_1$. Analogously, $z_1^{u}=z_1^{(1-p)^p}n_2$ for some $n_2\in N_2'$. Altogether,

$$\begin{aligned} t^{u}=(z_1^{-1}y_1)^{u}=z_1^{u}y_1^{u}= n_2^{-1}\cdot w^{-p\lambda }\cdot t\cdot v^{p\lambda }\cdot n_1, \end{aligned}$$

(5.8)

which belongs to M—here we replaced $z_1^{-(1-p)^p}=w^{-p\lambda }\cdot z_1^{-1}$ and $y_1^{(1-p)^p}=y_1\cdot v^{p^\lambda }$. Hence, $M\unlhd H$. Finally, by definition $H=M\cdot X^p$, and moreover

$$\begin{aligned} M\cap X^p\subseteq {{\,\textrm{Ker}\,}}(\theta )\cap X^p=\{1\}, \end{aligned}$$

so that $H=M\rtimes X^p$. This completes the proof of (ii).

Finally, by (i) and (ii) one has the isomorphism of p-elementary abelian groups

$$\begin{aligned} \begin{aligned} M/\Phi (M)&\simeq N_1/\Phi (N_1)\times N_2/\Phi (N_2)\times T/T^p \\ H/\Phi (H)&\simeq X^p/X^{p^2}\times M/M^p[M,H]. \end{aligned} \end{aligned}$$

(5.9)

From (5.8) one has that $[T,X^p]\subseteq \Phi (M)$, and since $H=MX^p$, $U_1=N_1X^p$, and $U_2=N_2X^p$, form (5.9) one deduces (iii). $\square$

5.3 The Subgroup H and Kummerianity

Proposition 5.3

The oriented pro-p group $(H,\theta \vert _H)$ is not Kummerian.

Proof

Let N be the normal subgroup of H generated as a normal subgroup by $N_1,N_2$, and set ${\bar{H}}=H/N$. Then $N\subseteq {{\,\textrm{Ker}\,}}(\theta \vert _H)$, and clearly ${\bar{H}}$ is finitely generated. Moreover, by duality the restriction map $\textrm{res}_{H,N}^1:H^1(H,\mathbb {Z}/p\mathbb {Z})\rightarrow H^1(N,\mathbb {Z}/p\mathbb {Z})^H$ is surjective, as by Lemma 5.2 one has

$$\begin{aligned} N/N^p[N,H]\simeq N_1/N_1^p[N_1,U_1]\times N_2/N_2^p[N_2,U_2], \end{aligned}$$

which embeds in $H/\Phi (H)$. In particular, $\{uN,tN\}$ is a minimal generating set of ${\bar{H}}$. Thus, by Proposition 2.10 if the oriented pro-p group $({\bar{H}},{\bar{\theta }})$ is not Kummerian—where ${\bar{\theta }}=(\theta \vert _H)_{/N}:{\bar{H}}\rightarrow 1+p\mathbb {Z}_p$ is the orientation induced by $\theta \vert _H$—then also $(H,\theta \vert _H)$ is not Kummerian.

By (5.8), in H one has that $[t,u^{-1}]\equiv 1\bmod N$, and thus ${\bar{H}}$ is abelian. Moreover,

$$\begin{aligned} {\bar{\theta }}(uN)=\theta (u)=(1-p)^p \qquad \text {and}\qquad {\bar{\theta }}(tN)=\theta (t)=1, \end{aligned}$$

so that ${{\,\textrm{Ker}\,}}({\bar{\theta }})=\langle tN\rangle$. Therefore, the subgroup $K_{{\bar{\theta }}}({\bar{H}})$ is generated by

$$\begin{aligned} \left( t^{-\theta (u)}utu^{-1}\right) N=t^{p^2\lambda }N. \end{aligned}$$

Thus, the quotient ${{\,\textrm{Ker}\,}}({\bar{\theta }})/K_{{\bar{\theta }}}({\bar{H}})=\langle tN\rangle /\langle tN\rangle ^{p^2}$ is not torsion-free, and by Proposition 2.2, $({\bar{H}},{\bar{\theta }})$ is not Kummerian. $\square$

This completes the proof of Theorem 1.1 case (1.1.b).

Remark 5.4

If $d_1=d_2=1$, case (1.1.b) of Theorem 1.1 is a particular case of [3, Proposition 6.5].

6 Massey Products

6.1 Massey Products in Galois Cohomology

Here we recall briefly what we need in order to prove Proposition 1.3. For a detailed account on Massey products for pro-p groups, we direct the reader to [8, 19, 35].

Let G be a pro-p group. For $n\ge 2$, the n-fold Massey product on $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ is a multi-valued map

$$\begin{aligned} \underbrace{\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})\times \cdots \times \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})}_{n\text { times}}\longrightarrow \textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z}). \end{aligned}$$

For $n\ge 2$, given a sequence $\alpha _1,\ldots ,\alpha _n$ of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ (with possibly $\alpha _i=\alpha _j$ for some $1\le i<j\le n$), the (possibly empty) subset of $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$ which is the value of the n-fold Massey product associated to the sequence $\alpha _1,\ldots ,\alpha _n$ is denoted by $\langle \alpha _1,\ldots ,\alpha _n\rangle$. If $n=2$, then the 2-fold Massey product coincides with the cup-product, i.e., for $\alpha _1,\alpha _2\in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ one has

$$\begin{aligned} \langle \alpha _1,\alpha _2\rangle =\{\alpha \smallsmile \alpha _2\}\subseteq \textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z}). \end{aligned}$$

(6.1)

A pro-p group G is said to satisfy:

• (a) the n -Massey vanishing property (with respect to $\mathbb {Z}/p\mathbb {Z}$) if for every sequence $\alpha _1,\ldots ,\alpha _n$ of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$, $\langle \alpha _1,\ldots ,\alpha _n\rangle \ne \varnothing$ implies $0\in \langle \alpha _1,\ldots ,\alpha _n\rangle$;

• (b) the strong n-Massey vanishing property (with respect to $\mathbb {Z}/p\mathbb {Z}$) if for every sequence $\alpha _1,\ldots ,\alpha _n$ of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$, the condition on the cup-products

  $$\begin{aligned} \alpha _1\smallsmile \alpha _2=\alpha _2\smallsmile \alpha _3=\cdots =\alpha _{n-1}\smallsmile \alpha _n=0 \end{aligned}$$

  (6.2)

  implies $0\in \langle \alpha _1,\ldots ,\alpha _n\rangle$ (cf. [21, Definition 1.2])—we remind that the triviality condition (6.2) is satisfied whenever $\langle \alpha _1,\ldots ,\alpha _n\rangle \ne \varnothing$, cf., e.g., [19, Sect. 2];

• (c) the cyclic p-Massey vanishing property if for every element $\alpha \in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$, the p-fold Massey product $\langle \alpha ,\ldots ,\alpha \rangle$ contains 0.

Remark 6.1

Given a sequence $\alpha _1,\ldots ,\alpha _n$ of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$, if an element $\omega$ of $\textrm{H}^2(G,\mathbb {Z}/p\mathbb {Z})$ is a value of the n-fold Massey product $\langle \alpha _1,\ldots ,\alpha _n\rangle$, then

$$\begin{aligned} \omega +\alpha _1\smallsmile \beta \in \langle \alpha _1,\ldots ,\alpha _n\rangle \qquad \text {and}\qquad \omega +\alpha _n\smallsmile \beta \in \langle \alpha _1,\ldots ,\alpha _n\rangle \end{aligned}$$

for any $\beta \in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ (cf. [19, Remark 2.2]).

In [18, Theorem 8.1], J. Minač and N.D. Tân proved that the maximal pro-p Galois group of a field $\mathbb {K}$ containing a root of 1 of order p (and also $\sqrt{-1}$ if $p=2$) satisfies the cyclic p-Massey vanishing property. The proof of the last property for a pro-p group G as in Theorem 1.1 is rather immediate.

Proof of Proposition 1.3–(ii)

By Proposition 4.1 and Proposition 5.1, G may complete into a Kummerian oriented pro-p group with torsion-free orientation. Hence, G satisfies the cyclic p-Massey vanishing property by [27, Theorem 3.10]. $\square$

6.2 Massey Products and Unipotent Upper-Triangular Matrices

Massey products for a pro-p group G may be translated in terms of unipotent upper-triangular representations of G as follows. For $n\ge 2$ let

$$\begin{aligned} \mathbb {U}_{n+1}=\left\{ \left( \begin{array}{ccccc} 1 &{} a_{1,2} &{} \cdots &{} &{} a_{1,n+1} \\ {} &{} 1 &{} a_{2,3} &{} \cdots &{} \\ &{}&{}\ddots &{}\ddots &{} \vdots \\ {} &{}&{}&{}1&{} a_{n,n+1} \\ {} &{}&{}&{}&{}1 \end{array}\right) \mid a_{i,j}\in \mathbb {Z}/p \right\} \subseteq \textrm{GL}_{n+1}(\mathbb {Z}/p\mathbb {Z}) \end{aligned}$$

be the group of unipotent upper-triangular $(n+1)\times (n+1)$-matrices over $\mathbb {Z}/p$. Then $\mathbb {U}_{n+1}$ is a finite p-group. Moreover, for $1\le h,l\le n+1$ let $E_{h,l}$ denote the $(n+1)\times (n+1)$ matrix with the (h, l)-entry equal to 1, and all the other entries equal to 0.

Now let $\rho :G\rightarrow \mathbb {U}_{n+1}$ be a homomorphism of pro-p groups. Observe that for every $h=1,\ldots ,n$, the projection $\rho _{h,h+1}:G\rightarrow \mathbb {Z}/p$ of $\rho$ onto the $(h,h+1)$-entry is a homomorphism, and thus it may be considered as an element of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$. One has the following “pro-p translation” of a result of W. Dwyer which interprets Massey product in terms of unipotent upper-triangular representations (cf., e.g., [11, Lemma 9.3]).

Proposition 6.2

Let G be a pro-p group, and let $\alpha _1,\ldots ,\alpha _n$ be a sequence of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$, with $n\ge 2$. Then the n-fold Massey product $\langle \alpha _1,\ldots ,\alpha _n\rangle$:

• (i) is not empty if, and only if, there exists a morphism of pro-p groups ${\bar{\rho }}:G\rightarrow \mathbb {U}_{n+1}/\textrm{Z}(\mathbb {U}_{n+1})$ such that ${\bar{\rho }}_{h,h+1}=\alpha _h$ for every $h=1,\ldots ,n$;

• (ii) vanishes if, and only if, there exists a morphism of pro-p groups $\rho :G\rightarrow \mathbb {U}_{n+1}$ such that $\rho _{h,h+1}=\alpha _h$ for every $h=1,\ldots ,n$.

We recall that

$$\begin{aligned} \textrm{Z}(\mathbb {U}_{n+1})=\left\{ \,I_{n+1}+aE_{1,n+1}\,\mid \,a\in \mathbb {Z}/p\mathbb {Z}\,\right\} \simeq \mathbb {Z}/p\mathbb {Z}. \end{aligned}$$

We use this fact to prove statements (iii.a)–(iii.b) of Proposition 1.3. First of all, let G be as in Theorem 1.1, and let $\alpha _1,\ldots ,\alpha _n$ be a sequence of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$. Keeping the same notation as in Sect. 3.3, for $h=1,\ldots ,n$ one has

$$\begin{aligned} \alpha _h=\alpha _h(x)\cdot \chi +\sum _{i=1}^{d_1}\alpha _h(y_i)\cdot \varphi _i+\sum _{j=1}^{d_2}\alpha _h(z_j)\cdot \psi _j. \end{aligned}$$

Therefore, for $h=1,\ldots ,n-1$ one obtains

$$\begin{aligned} \alpha _h\smallsmile \alpha _h=S_h\cdot (\chi \smallsmile \varphi _1)+S'_h\cdot (\chi \smallsmile \psi _1), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} S_h=&(\alpha _h(x)\alpha _{h+1}(y_1)-\alpha _{h}(y_1)\alpha _{h+1}(x))\\&+(-1)^\epsilon \sum _{2\mid i} (\alpha _h(y_i)\alpha _{h+1}(y_{i+1})-\alpha _{h}(y_{i+1})\alpha _{h+1}(y_i)), \\ S'_h =&(\alpha _h(x)\alpha _{h+1}(z_1)-\alpha _{h}(z_1)\alpha _{h+1}(x))\\&+(-1)^{\epsilon }\sum _{ 2\mid j} (\alpha _h(z_j)\alpha _{h+1}(z_{j+1})-\alpha _{h}(z_{j+1})\alpha _{h+1}(z_j)), \end{aligned}\end{aligned}$$

with $\epsilon =0$ if G is as in (1.1.a), and $\epsilon =1$ if G is as in (1.1.b). If the sequence $\alpha _1,\ldots ,\alpha _n$ satisfies condition (6.2), then one has $S_h=S'_h=0$ for $h=1,\ldots ,n-1$, as $\{\chi \smallsmile \varphi _1,\chi \smallsmile \psi _1\}$ is a basis of $\textrm{H}^2(G,\mathbb {Z}/p)$.

From now on, we will assume that $p>3$ while considering a pro-p group G as in (1.1.b), unless stated otherwise.

6.3 3-Fold Massey Products

We are ready to prove the following.

Proposition 6.3

A pro-p group G satisfies the 3-Massey vanishing property in the following cases:

• (a) if G is as in (1.1.a);

• (b) if G is as in (1.1.b) and $p>3$.

Proof

Let $\alpha _1,\alpha _2,\alpha _3$ be a sequence of elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ satisfying (6.2). Then $S_1=S_1'=S_2=S_2'=0$ (cf. Sect. 6.2). Our goal is to construct a morphism $\rho :G\rightarrow \mathbb {U}_4$ such that $\rho _{1,2}=\alpha _1$, $\rho _{2,3}=\alpha _2$, $\rho _{3,4}=\alpha _3$.

For every $w\in \mathcal {X}$ set

$$\begin{aligned} A(w)=I+\alpha _1(w)E_{1,2}+\alpha _2(w)E_{2,3}+\alpha _3(w)E_{3,4}\in \mathbb {U}_4, \end{aligned}$$

where I denotes the $4\times 4$ identity matrix. If G is as in (1.1.a), then one computes

$$\begin{aligned} \begin{aligned} C&=[A(x),A(y_1)]\cdots [A(y_{d_1-1}),A(y_{d_1})] \\ {}&= I+E_{1,4}\left( \alpha _1(y_1)\alpha _2(x)\alpha _3(y_1)+\sum _{2\mid i}\alpha _1(y_i)\alpha _2(y_{i+1})\alpha _3(y_i)\right) \\ C'&=[A(x),A(z_1)]\cdots [A(z_{d_2-1}),A(z_{d_2})] \\ {}&= I+E_{1,4}\left( \alpha _1(z_1)\alpha _2(x)\alpha _3(z_1)+\sum _{2\mid j}\alpha _1(z_j)\alpha _2(z_{j+1})\alpha _3(z_j)\right) ; \end{aligned}\end{aligned}$$

(6.3)

while if G is as in (1.1.b), then one computes

$$\begin{aligned} \begin{aligned} C&=A(y_1)^p[A(y_1),A(x)]\cdots [A(y_{d_1-1}),A(y_{d_1})] \\&= I+E_{1,4}\left( \alpha _1(x)\alpha _2(y_1)\alpha _3(x)+\sum _{2\mid i}\alpha _1(y_i)\alpha _2(y_{i+1})\alpha _3(y_i)\right) \\ C'&=A(z_1)^p[A(z_1),A(x)]\cdots [A(z_{d_2-1}),A(z_{d_2})] \\&= I+E_{1,4}\left( \alpha _1(x)\alpha _2(z_1)\alpha _3(x)+\sum _{2\mid j}\alpha _1(z_j)\alpha _2(z_{j+1})\alpha _3(z_j)\right) . \end{aligned}\end{aligned}$$

(6.4)

—observe that the exponent of $\mathbb {U}_4$ is p, as $p>4$, and thus $A(y_1)^p=A(z_1)^p=I$.

In both cases, $C,C'\in \textrm{Z}(\mathbb {U}_4)$, and therefore the assignment $w\mapsto A(w)$ for every $w\in \mathcal {X}$ yields a morphism ${\bar{\rho }}:G\rightarrow \mathbb {U}_4/\textrm{Z}(\mathbb {U}_4)$ satisfying ${\bar{\rho }}_{h,h+1}=\alpha _h$ for $h=1,2,3$. Thus, $\langle \alpha _1,\alpha _2,\alpha _3\rangle \ne \varnothing$ by Proposition 6.2.

Moreover, if $C=C'=I$ then the same assignment yields a morphism $\rho :G\rightarrow \mathbb {U}_4$ with the desired properties. In particular, by (6.3)–(6.4) one has $C=I$ if $\alpha _1(w)=\alpha _3(w)=0$ for every $w=y_1,\ldots ,y_{d_1}$, or for every $w=y_2,\ldots ,y_{d_1}$ and $w=x$; and analogously $C'=I$ if $\alpha _1(w)=\alpha _3(w)=0$ for every $w=z_1,\ldots ,z_{d_{d_2}}$, or for every $w=z_2,\ldots ,z_{d_2}$ and $w=x$.

On the other hand, if $C\ne I$ then $\chi \smallsmile \varphi _1=\pm {{\,\textrm{trg}\,}}(r_1 G_{(3)})$ belongs to $\langle \alpha _1,\alpha _2,\alpha _3\rangle$, and analogously if $C'\ne I$ then $\chi \smallsmile \psi _1=\pm {{\,\textrm{trg}\,}}(r_2 G_{(3)})$ belongs to $\langle \alpha _1,\alpha _2,\alpha _3\rangle$ (cf. [19, Lemma 3.7])—here the sign depends on whether the relations are as in (1.1.a) or in (1.1.b). Now, if $\alpha _h(y_i)\ne 0$ for some $h=1,3$ and $i\in \{2,\ldots ,d_1\}$, then

$$\begin{aligned} \chi \smallsmile \varphi _1=\alpha _h\smallsmile \beta \qquad \text {for some }\beta \in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}). \end{aligned}$$

Analogously, if $\alpha _h(z_j)\ne 0$ for some $h=1,3$ and $j\in \{2,\ldots ,d_2\}$, then

$$\begin{aligned} \chi \smallsmile \psi _1=\alpha _h\smallsmile \beta \qquad \text {for some }\beta \in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z}). \end{aligned}$$

Moreover, if $\alpha _h(x)\ne 0$ for some $h=1,3$, then

$$\begin{aligned} \chi \smallsmile \varphi _1=\alpha _h\smallsmile \beta \qquad \text {and}\qquad \chi \smallsmile \psi _1=\alpha _h\smallsmile \beta ' \end{aligned}$$

for some $\beta ,\beta '\in \textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$. Therefore, Remark 6.1 implies that if $C\ne I$ or $C'\ne I$ then $0\in \langle \alpha _1,\alpha _2,\alpha _3\rangle$ anyway. $\square$

Remark 6.4

If $p=3$ and G as in (1.1.b), then G does not satisfy the 3-Massey vanishing property. Indeed, set $\alpha _1=\alpha _3=\varphi _1+\psi _1$, and $\alpha _2=\varphi _1$. Then

$$\begin{aligned} \alpha _1\smallsmile \alpha _2=\alpha _2\smallsmile \alpha _3=\pm (\varphi _1\smallsmile \psi _1)=0. \end{aligned}$$

It is easy to see that one may construct a morphism of pro-p groups ${\bar{\rho }}:G\rightarrow \mathbb {U}_4/\textrm{Z}(\mathbb {U}_4)$ such that ${\bar{\rho }}_{1,2}={\bar{\rho }}_{3,4}=\alpha _1$ and ${\bar{\rho }}_{2,3}=\alpha _2$—and thus $\langle \alpha _1,\alpha _2,\alpha _1\rangle \ne \varnothing$ by Proposition 6.2—but, on the other hand, one may not construct a morphism of pro-p groups $\rho :G\rightarrow \mathbb {U}_4$ satisfying $\rho _{1,2}=\rho _{3,4}=\alpha _1$ and $\rho _{2,3}=\alpha _2$—so that $0\notin \langle \alpha _1,\alpha _2,\alpha _1\rangle$ by Proposition 6.2.

6.4 4-Fold Massey Products

Proposition 6.5

A pro-p group G as in Theorem 1.1 satisfies the strong 4-Massey vanishing property.

Proof

Let $\alpha _1,\ldots ,\alpha _4$ be a sequence of four elements of $\textrm{H}^1(G,\mathbb {Z}/p\mathbb {Z})$ satisfying (6.2). Our goal is to construct a homomorphism of pro-p groups $\rho :G\rightarrow \mathbb {U}_5$ such that $\rho _{h,h+1}=\alpha _h$ for $h=1,\ldots ,5$, so that the claim follows by Proposition 6.2.

Let I denote the identity matrix of the group $\mathbb {U}_5$. For every $w\in \mathcal {X}=\{x,y_1,\ldots ,z_{d_2}\}$ set

$$\begin{aligned} A(w)=\left( \begin{array}{ccccc} 1 &{}\alpha _1(w)&{}0&{}0&{}0 \\ &{}1&{}\alpha _2(w)&{}0&{}0 \\ {} &{}&{}1&{}\alpha _3(w)&{}0 \\ {} &{}&{}&{}1&{}\alpha _4(w) \\ {} &{}&{}&{}&{}1 \end{array} \right) \in \mathbb {U}_5. \end{aligned}$$

Moreover, put

$$\begin{aligned}\begin{aligned} C&=(c_{hl})= A(y_1)^{\epsilon p}\cdot [A(x),A(y_1)]^{(-1)^\epsilon }\cdots \left[ A(y_{d_1-1}),A(y_{d_1})\right] ,\\ C'&=(c'_{hl})= A(z_1)^{\epsilon p}\cdot [A(x),A(z_1)]^{(-1)^\epsilon }\cdots \left[ A(z_{d_2-1}),A(z_{d_2})\right] . \end{aligned}\end{aligned}$$

We will consider the matrix C as a function of the matrices $A(x),\ldots ,A(y_{d_1})$, and the matrix $C'$ as a function of the matrices $A(x),A(z_1),\ldots ,A(z_{d_2})$.

Since $p\ge 5$, the exponent of the p-group $\mathbb {U}_5$ is p, and thus $A(y_1)^p=A(z_1)^p=I$. Moreover, for every $w,w'\in \mathcal {X}$, the $(h,h+1)$-entry of $[A(w),A(w')]$ is 0 for every $h=1,\ldots ,4$, and thus also $c_{h,h+1}=c'_{h,h+1}=0$. Moreover, for $h=1,2,3$ one has $c_{h,h+2}=S_h$ and $c'_{h,h+2}=S'_h$—which are equal to 0 by (6.2).

We split the proof in the analysis of the following three cases. Our aim is to modify suitably the matrices A(w)—without modifying the $(h,h+1)$-entries with $h=1,\ldots ,4$—to obtain $C=C'=I$.

Case 1. Suppose first that:

• (1.a) $\alpha _2(x)=\alpha _2(y_i)=0$ for all $2\le i\le d_1$; or

• (1.b) $\alpha _3(x)=\alpha _3(y_i)=0$ for all $2\le i\le d_1$.

Since $S_1=S_2=S_3=0$ by (6.2), one has

$$\begin{aligned} \alpha _1(x)\alpha _2(y_1)=\alpha _2(y_1)\alpha _3(x)= & {} 0, \end{aligned}$$

(6.5)

$$\begin{aligned} \alpha _2(x)\alpha _3(y_1)=\alpha _3(y_1)\alpha _4(x)= & {} 0, \end{aligned}$$

(6.6)

respectively, in case (1.a) and in case (1.b). Applying (6.5)–(6.6), one computes

$$\begin{aligned}{}[A(y_1),A(x)]={\left\{ \begin{array}{ll} I+\left( \alpha _3(y_1)\alpha _4(x)-\alpha _3(x)\alpha _4(y_1)\right) E_{3,5} &{}\quad \text {in case (1.a)}, \\ I+\left( \alpha _1(y_1)\alpha _2(x)-\alpha _2(x)\alpha _1(y_1)\right) E_{1,3} &{}\quad \text {in case (1.b)}, \end{array}\right. }\end{aligned}$$

and

$$\begin{aligned}{} & {} [A(y_i),A(y_{i+1})]=\\{} & {} \quad {\left\{ \begin{array}{ll} I+\left( \alpha _3(y_i)\alpha _4(y_{i+1})-\alpha _3(y_{i+1})\alpha _4(y_i)\right) E_{3,5} &{}\quad \text {in case (1.a)}, \\ I+\left( \alpha _1(y_i)\alpha _2(y_{i+1})-\alpha _2(y_{i+1})\alpha _1(y_i)\right) E_{1,3} &{}\quad \text {in case (1.b)}, \end{array}\right. }\end{aligned}$$

for $i=2,4,\ldots ,d_1-1$. Altogether, one has $C=I+S_3E_{3,5}$ in case (1.a) and $C=I+S_1E_{1,3}$ in case (1.b), so that in both cases $C=I$ by (6.2).

Analogously, if $\alpha _2(x)=\alpha _2(z_j)=0$ for all $2\le j\le d_2$, or if $\alpha _3(x)=\alpha _3(z_j)=0$ for all $2\le j\le d_2$, then $C'=I$. This completes the analysis of case 1.

Case 2. Now suppose that $\alpha _1(x)=\alpha _4(x)=\alpha _1(y_i)=\alpha _4(y_i)=0$ for all $2\le i\le d_1$. Since $S_1=S_2=S_3=0$ by (6.2), one has

$$\begin{aligned} \alpha _1(y_1)\alpha _2(x)=\alpha _3(x)\alpha _4(y_1)=0. \end{aligned}$$

(6.7)

Then one computes

$$\begin{aligned} \begin{aligned}&[A(y_1),A(x)] = I+\left( \alpha _2(y_1)\alpha _3(x)-\alpha _2(x)\alpha _3(y_1)\right) E_{2,4}+\alpha _2(x)\alpha _3(y_1)\alpha _4(y_1)E_{2,5},\\&[A(y_i),A(y_{i+1})] =I+\left( \alpha _2(y_i)\alpha _3(y_{i+1})-\alpha _2(y_{i+1})\alpha _3(y_i)\right) E_{2,4}, \end{aligned} \end{aligned}$$

where we apply (6.7) to obtain the first equality, and in the second one i runs through the even positive integers between 2 and $d_1-1$. If $\alpha _2(x)\alpha _3(y_1)\alpha _4(y_1)=0$ then it is straightforward to see that $C=I+S_2E_{2,4}=I$. Otherwise, $\alpha _2(x)\ne 0$, so that (6.7) implies that $\alpha _1(y_1)=0$. In this case, set

$$\begin{aligned} {\tilde{A}}=I-\alpha _3(y_1)\alpha _4(y_1)E_{3,5}. \end{aligned}$$

Then

$$\begin{aligned} \left[ \tilde{A},A(x)\right] =I-\alpha _2(x)\alpha _3(y_1)\alpha _4(y_1)E_{2,5}, \end{aligned}$$

and

$$\begin{aligned}\begin{aligned} \left[ A(y_1){\tilde{A}},A(x)\right]&= \underbrace{\left[ A(y_1),[{\tilde{A}},A(x)]\right] }_{=I}\left[ {\tilde{A}},A(x)\right] [A(y_1),A(x)]\\&= I+\left( \alpha _2(y_1)\alpha _3(x)-\alpha _2(x)\alpha _3(y_1)\right) E_{2,4}. \end{aligned} \end{aligned}$$

Therefore, replacing $A(y_1)$ with $A(y_1){\tilde{A}}$ yields $c_{2,4}=S_2=0$ and $C_{hl}=0$ for $h<l$, i.e., $C=I$.

An analogous argument yields $C'=I$—after replacing suitably the matrix $A(z_1)$ if needed—if $\alpha _1(x)=\alpha _3(x)=\alpha _1(z_j)=\alpha _3(z_j)=0$ for all $1\le j\le d_2$. This completes the analysis of case 2.

Case 3. Finally, if none of the above two assumptions on the triviality of the values $\alpha _h(x)$ and $\alpha _h(y_i)$, with $2\le i\le d_1$, hold true, then

• (3.a) there are $w,w'\in \{x,y_2,\ldots ,y_{d_1}\}$—possibly $w=w'$—such that $\alpha _1(w)\ne 0$ and $\alpha _2(w')\ne 0$, or

• (3.b) there are $w,w'\in \{x,y_2,\ldots ,y_{d_1}\}$—possibly $w=w'$—such that $\alpha _4(w)\ne 0$ and $\alpha _3(w')\ne 0$.

Suppose we are in case (3.a). If $w=x$ or $w=y_i$ with i odd, set

$$\begin{aligned} {\tilde{A}}={\left\{ \begin{array}{ll} I+\frac{c_{1,4}}{\alpha _1(w)}E_{2,4}&{}\quad \text {if }w\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I-\frac{c_{1,4}}{\alpha _1(w)}E_{2,4}&{} \quad \text {if }w\in \{\,y_i\,\mid \,i\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}$, if $w=y$ with i even. After the replacement, one has $c_{hl}=0$ for $h< l\le h+2$, and for $(h,l)=(1,4)$. Then, set

$$\begin{aligned} {\tilde{A}}'={\left\{ \begin{array}{ll} I+\frac{c_{2,5}}{\alpha _1(w')}E_{3,5}&{} \quad \text {if }w'\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I-\frac{c_{2,5}}{\alpha _1(w')}E_{3,5}&{} \quad \text {if }w'\in \{\,y_i\,\mid \,\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}'$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}'$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}'$, if $w=y$ with i even. After this further replacement, one has $c_{hl}=0$ for $h< l\le h+3$. Finally, set

$$\begin{aligned} {\tilde{A}}''={\left\{ \begin{array}{ll} I+\frac{c_{1,5}}{\alpha _1(w)}E_{2,5}&{} \quad \text {if }w\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I-\frac{c_{1,5}}{\alpha _1(w)}E_{2,5}&{} \quad \text {if }w\in \{\,y_i\,\mid \,i\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}''$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}''$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}''$, if $w=y$ with i even. After this last replacement, one has $C=I$.

Now suppose we are in case (3.b). If $w=x$ or $w=y_i$ with i odd, set

$$\begin{aligned} {\tilde{A}}={\left\{ \begin{array}{ll} I-\frac{c_{2,5}}{\alpha _4(w)}E_{3,4}&{} \quad \text {if }w\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I+\frac{c_{2,5}}{\alpha _4(w)}E_{3,4}&{} \quad \text {if }w\in \{\,y_i\,\mid \, i\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}$, if $w=y$ with i even. After the replacement, one has $c_{hl}=0$ for $h< l\le h+2$, and for $(h,l)=(2,5)$. Then, set

$$\begin{aligned} {\tilde{A}}'={\left\{ \begin{array}{ll} I-\frac{c_{1,4}}{\alpha _3(w')}E_{1,3}&{} \quad \text {if }w'\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I+\frac{c_{1,4}}{\alpha _3(w')}E_{1,3}&{} \quad \text {if }w'\in \{\,y_i\,\mid \,i\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}'$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}'$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}'$, if $w=y$ with i even. After this further replacement, one has $c_{hl}=0$ for $h< l\le h+3$. Finally, set

$$\begin{aligned} {\tilde{A}}''={\left\{ \begin{array}{ll} I-\frac{c_{1,5}}{\alpha _1(w)}E_{1,4}&{} \quad \text {if }w\in \{\,x,y_3,\ldots ,y_{d_1}\,\}\\ I+\frac{c_{1,5}}{\alpha _1(w)}E_{1,4}&{} \quad \text {if }w\in \{\,y_i\,\mid \, i\text { is even}\,\}, \end{array}\right. } \end{aligned}$$

and replace $A(y_1)$ with $A(y_1){\tilde{A}}''$, if $w=x$, or $A(y_{i-1})$ with $A(y_{i-1}){\tilde{A}}''$ if $w=y_i$ with i odd, or $A(y_{i+1})$ with $A(y_{i+1}){\tilde{A}}''$, if $w=y$ with i even. After this last replacement, one has $C=I$.

Moreover, if none of the above two assumptions on the triviality of the values $\alpha _h(x)$ and $\alpha _h(z_j)$, with $2\le j\le d_2$, hold true, the same argument produces suitable matrices $A(z_1),\ldots ,A(z_{d_2})$ such that the matrix $C'$ is the identity matrix. This concludes the analysis of case 3.

Altogether, the assignment $w\mapsto A(x)$ for every $w\in \mathcal {X}$—with the matrices A(w)’s suitably modified in case of need—yields a homomorphism of pro-p groups $\rho :G\rightarrow \mathbb {U}_5$ with the desired properties. $\square$

We believe that the answer to the following questions is positive.

Question 6.6

• (a) Let G be as in (1.1.a). Does G satisfy the strong n-Massey vanishing property for every $n\ge 3$?

• (b) Let G be as in (1.1.b). Does G satisfy the strong n-Massey vanishing property for every $3\le n<p$?

7 The Minač–Tân Pro-p Group

We focus now on the Minač–Tân pro-p group

$$\begin{aligned} G=\langle \,x_1,\ldots ,x_5\,\mid \,r=1\,\rangle \qquad \text {with }r=\left[ [x_1,x_2],x_3\right] [x_4,x_5]. \end{aligned}$$

Using Proposition 6.2, one may show that G does not satisfy the 3-Massey vanishing property (cf. [19, Example 7.2]). Our aim is to show that G cannot complete into a 1-cyclotomic oriented pro-p group with torsion-free orientation.

7.1 Kummerianity and 1-Cyclotomicity

Proposition 7.1

Let G be the Minač–Tân pro-p group, and let $\theta :G\rightarrow 1+p\mathbb {Z}_p$ be a torsion-free orientation. Then the oriented pro-p group $(G,\theta )$ is Kummerian if, and only if, $x_4,x_5\in {{\,\textrm{Ker}\,}}(\theta )$, and:

• (a) $x_3\in {{\,\textrm{Ker}\,}}(\theta )$; or

• (b) $x_1,x_2\in {{\,\textrm{Ker}\,}}(\theta )$.

Proof

Let $c:G\rightarrow \mathbb {Z}_p(\theta )$ be an arbitrary continuous 1-cocycle, and set $c(x_i)=\lambda _i$ for $i=1,\ldots ,5$. Applying (2.2)–(2.3) one computes $c(r)=c([[x_1,x_2],x_3])+c([x_4,x_5])$, and

$$\begin{aligned} c([[x_1,x_2],x_3])= & {} \theta (x_1x_2)^{-1}\left( \theta (x_3)^{-1}-1\right) \left( \lambda _1(1-\theta (x_2))-\lambda _2(1-\theta (x_1))\right) ,\nonumber \\ c([x_4,x_5])= & {} \theta (x_4x_5)^{-1}\left( \lambda _4(1-\theta (x_5))-\lambda _5(1-\theta (x_4))\right) . \end{aligned}$$

(7.1)

On the other hand, $c(r)=0$ as $r=1$.

Suppose that $(G,\theta )$ is Kummerian. Then by Lemma 2.9, we may prescribe arbitrary values to $\lambda _1,\ldots ,\lambda _5$. If $\lambda _4=1$ and $\lambda _i=0$ for $i\ne 4$, from (7.1) and from the fact that $c(r)=0$ one obtains $0=1\cdot (1-\theta (x_5))$, and thus $\theta (x_5)=1$. Analogously, if $\lambda _5=1$ and $\lambda _i=0$ for $i\ne 5$, one deduces $\theta (x_4)=1$. Finally, if $\lambda _4=\lambda _5=0$ from (7.1) one obtains

$$\begin{aligned} 0=c(r)=\theta (x_1x_2)^{-1}\left( \theta (x_3)^{-1}-1\right) \left( \lambda _1(1-\theta (x_2))-\lambda _2(1-\theta (x_1))\right) , \end{aligned}$$

and the arbitrariness of $\lambda _1,\lambda _2$ implies that $\theta (x_3)=1$ or $\theta (x_1)=\theta (x_2)=1$.

Conversely, suppose that $x_4,x_5\in {{\,\textrm{Ker}\,}}(\theta )$, and at least one of the hypothesis (i)–(ii) holds true. Then for any choice for $\lambda _4,\lambda _5$, by (7.1) one has $c([x_4,x_5])=0$. On the other hand, one has

$$\begin{aligned} c([[x_1,x_2],x_3])={\left\{ \begin{array}{ll} 0\cdot (\lambda _1(1-\theta (x_2))-\lambda _2(1-\theta (x_1)))=0 &{}\quad \text {if }x_3\in {{\,\textrm{Ker}\,}}(\theta ), \\ \left( \theta (x_3)^{-1}-1\right) (\lambda _1\cdot 0-\lambda _2\cdot 0)=0 &{}\quad \text {if }x_1,x_2\in {{\,\textrm{Ker}\,}}(\theta ). \end{array}\right. }\end{aligned}$$

Altogether, any choice for $\lambda _1,\ldots ,\lambda _5$ yields a well-defined continuous 1-cocycle $c:G\rightarrow \mathbb {Z}_p(\theta )$, and thus $(G,\theta )$ is Kummerian by Lemma 2.9. $\square$

Now consider the subgroup H of G generated by $x_3,x_4,x_5$ and by $y=[x_1,x_2]$. Then H is subject to the relation

$$\begin{aligned} r=[y,x_3][x_4,x_5]=1. \end{aligned}$$

If $(G,\theta )$ is a 1-cyclotomic oriented pro-p group, with $\theta$ a torsion-free orientation, then $=(H,\theta \vert _H)$ is Kummerian. Therefore, if $c':H\rightarrow \mathbb {Z}_p(\theta \vert _H)$ is a continuous 1-cocycle, applying (2.2)–(2.3) one obtains

$$\begin{aligned} \begin{aligned} 0=c'(r)&=c'([y,x_3])+c'([x_4,x_5])\\ {}&=\theta (yx_3)^{-1}\left( c'(y)(1-\theta (x_3))-c'(x_3)(1-\theta (y))\right) +0\\&=\theta (yx_3)^{-1}c'(y)(1-\theta (x_3)), \end{aligned} \end{aligned}$$

as $\theta (x_4)=\theta (x_5)=1$ by Proposition 7.1, and $y\in G'\subseteq {{\,\textrm{Ker}\,}}(\theta )$. Since $c'(y)$ may be arbitrarily chosen by Lemma 2.9, one deduces $\theta (x_3)=1$. This proves the following.

Lemma 7.2

Let G be the Minač–Tân pro-p group, and let $\theta :G\rightarrow 1+p\mathbb {Z}_p$ be a torsion-free orientation. If the oriented pro-p group $(G,\theta )$ is 1-cyclotomic then $x_3,x_4,x_5\in {{\,\textrm{Ker}\,}}(\theta )$.

Moreover, if $(G,\theta )$ is 1-cyclotomic we may suppose without loss of generality that $x_2\in {{\,\textrm{Ker}\,}}(\theta )$, too. Indeed, let $v_p:\mathbb {Z}_p\rightarrow \mathbb {N}$ denote the p-adic valuation, and let $k\ge 1$ be such that ${{\,\textrm{Im}\,}}(\theta )=1+p^k\mathbb {Z}_p$.

Suppose first that $v_p(\theta (x_2)-1)=k$ and $v_p(\theta (x_1)-1)>k$, and set $z=x_2x_1$. Then $\{z,x_2,x_3,x_4,x_5\}$ is a minimal generating set of G, $v_p(\theta (z)-1)=k$, and G is subject to the relation

$$\begin{aligned} \left[ [z,x_2],x_3\right] [x_4,x_5]=1, \end{aligned}$$

as $[x_2x_1,x_2]=[x_1,x_2]$. Hence, we may assume $v_p(\theta (x_1)-1)=k$.

Consequently, there exists $\lambda \in \mathbb {Z}_p$ such that $\theta (x_2)=\theta (x_1)^\lambda$. Now set $z=x_1^{-\lambda }x_2$. Then $\{x_1,z,x_3,x_4,x_5\}$ is a minimal generating set of G, $\theta (z)=\theta (x_2)\theta (x_1)^{-\lambda }=1$, and G is subject to the relation

$$\begin{aligned} \left[ [x_1,z],x_3\right] [x_4,x_5]=1, \end{aligned}$$

as $[x_1,x_1^{-\lambda } x_2]=[x_1,x_2]$.

Therefore, from now on $\theta :G\rightarrow 1+p\mathbb {Z}_p$ will denote a torsion-free orientation satisfying $x_2,\ldots ,x_5\in {{\,\textrm{Ker}\,}}(\theta )$.

7.2 The Subgroup U

Put $u=x_1^p$ and $t=x_1^{-1}x_3$. Let $\phi :G\rightarrow \mathbb {Z}/p$ be the homomorphism defined by $\phi (x_1)=\phi (x_3)=1$ and $\phi (x_i)=0$ for $i=2,4,5$, and let U be the kernel of $\phi$. Then U is a normal subgroup of G of index p, and it is generated as a normal subgroup of G by $\{u,t,x_2,x_4,x_5\}$. In fact, U is generated as a pro-p group by the set

$$\begin{aligned} \mathcal {X}_U=\left\{ \,u,\, t^{x_1^h},\,x_2^{x_1^h},\,x_4^{x_1^h},\,x_5^{x_1^h}\,\mid \,h=0,\ldots ,p-1\,\right\} , \end{aligned}$$

as $G/U=\{U,x_1U,\ldots ,x_1^{p-1}U\}$. We need to find a subset of $\mathcal {X}_U$ which minimally generates U as a pro-p group.

Proposition 7.3

The set

$$\begin{aligned} \mathcal {Y}_U=\left\{ \,t,\,x_2,\,x_2^{x_1},\, t^{x_1^h},\,x_4^{x_1^h},\,x_5^{x_1^h}\,\mid \,h=0,\ldots ,p-1\,\right\} , \end{aligned}$$

is a minimal generating set of U as a pro-p group. Moreover, the abelian pro-p group $U^{{{\,\textrm{ab}\,}}}$ is not torsion-free.

Proof

The subgroup U is the pro-p group generated by $\mathcal {X}_U$ and subject to the p-relations $r^{x_1^h}=1$, $h=0,\ldots ,p-1$. Since $x_3=x_1t$, one computes

$$\begin{aligned} \begin{aligned} {[}[x_1,x_2],x_3]&= [x_1,x_2]^{-1}\cdot [x_1,x_2]^{x_3}\\&=[x_2,x_1]\cdot \left[ x_1,x_2^{x_1}\right] ^{t}\\&=x_2^{-1}\cdot x_2^{x_1}\cdot \left( \left( x_2^{x_1^2}\right) ^{-1}x_2^{x_1}\right) ^{t}. \end{aligned} \end{aligned}$$

(7.2)

From (7.2), and from the relation $r=1$, one deduces the equivalence

$$\begin{aligned} \left( x_2^{x_1^2}\right) ^{-1}\cdot \left( x_2^{x_1}\right) ^2\cdot x_1^{-1}\equiv 1\mod U', \end{aligned}$$

(7.3)

as $[x_4,x_5]\in U'$ and $t\in U$.

Hence, $U^{{{\,\textrm{ab}\,}}}$ is the abelian pro-p group generated by $\mathcal {X}_{U^{{{\,\textrm{ab}\,}}}}=\{wU'\,\mid \,w\in \mathcal {X}_U\}$ and subject to the p relations induced by the equivalences $((x_2^{x_1^2})^{-1}(x_2^{x_1})^2 x_1^{-1})^{x_1^h}\equiv 1\bmod U'$, namely

$$\begin{aligned} \begin{aligned} x_2^{x_1^2}&\equiv \left( x_2^{x_1}\right) ^2 x_1^{-1}\mod U',\qquad \text {for }h=0,\\ x_2^{x_1^3}&\equiv \left( x_2^{x_1^2}\right) ^2 \left( x_1^{x_2}\right) ^{-1}\equiv \left( x_2^{x_1}\right) ^3 x_1^{-2}\mod U',\qquad \text {for }h=1,\\&\quad \vdots \\ x_2^{x_1^{p}}&\equiv \left( x_2^{x_1^{p-1}}\right) ^2 \left( x_1^{p-2}\right) ^{-1}\equiv \left( x_2^{x_1}\right) ^p x_1^{1-p}\mod U',\qquad \text {for }h=p-2,\\ x_2^{x_1^{p+1}}&\equiv \left( x_2^{x_1}\right) ^2\cdot x_1^{-1}\equiv \left( x_2^{x_1}\right) ^{p+1} x_1^{-p}\mod U',\qquad \text {for }h=p-1. \end{aligned} \end{aligned}$$

(7.4)

On the one hand, from (7.4) one deduces that the coset $x_2^{x_1^h}U'$ is generated by $x_2U'$ and $x_2^{x_1}U'$ for every $h=2,\ldots ,p-1$, so that $\mathcal {Y}_{U^{{{\,\textrm{ab}\,}}}}=\{wU'\,\mid \,w\in \mathcal {Y}_U\}$ generates $U^{{{\,\textrm{ab}\,}}}$ as an abelian pro-p group. On the other hand, from the equivalences with $h=p-2$ and $h=p-1$ one deduces that

$$\begin{aligned}\begin{aligned} \left( x_2^{x_1}\right) ^p x_1^{1-p}\left( x_2^{u}\right) ^{-1}&\equiv \left( x_2^{x_1}\right) ^p x_1^{1-p-1}\equiv \left( x_2^{x_1} x_1^{-1}\right) ^p\equiv 1\mod U',\\ \left( x_2^{x_1}\right) ^{p+1} x_1^{-p}\left( x_2^{ux_1}\right) ^{-1}&\equiv \left( x_2^{x_1}\right) ^{p+1-1} x_1^{-p}\equiv \left( x_2^{x_1} x_1^{-1}\right) ^p\equiv 1\mod U', \end{aligned} \end{aligned}$$

as $x_2^u\equiv x_2\bmod U'$; therefore, they yield equivalent relations in $U^{{{\,\textrm{ab}\,}}}$. Altogether, $U^{{{\,\textrm{ab}\,}}}$ is the abelian pro-p group minimally generated by $\mathcal {X}_{U^{{{\,\textrm{ab}\,}}}}$ and subject to the relation

$$\begin{aligned} \left( (x_2U')^{-1}\cdot x_2^{x_1}U'\right) ^p=1. \end{aligned}$$

Hence, $U^{{{\,\textrm{ab}\,}}}$ is not torsion-free, and $\mathcal {Y}_U$ is a minimal generating set of U by Fact 2.1. $\square$

From Proposition 7.3, one deduces that G is not absolutely torsion-free, and thus the oriented pro-p group $(G,\textbf{1})$ is not 1-cyclotomic.

7.3 1-Cyclotomicity and the Minač–Tân Pro-p Group

We are ready to prove Theorem 1.4.

Proof

Suppose for contradiction that there exists a torsion-free orientation $\theta :G\rightarrow 1+p\mathbb {Z}_p$ such that the oriented pro-p group $(G,\theta )$ is 1-cyclotomic. Then by Sect. 7.1, we may assume without loss of generality that $x_2,\ldots ,x_5\in {{\,\textrm{Ker}\,}}(\theta )$, while $\theta (x_1)\ne 1$ by Sect. 7.2. Set $\lambda \in p\mathbb {Z}_p\smallsetminus \{0\}$ such that $\theta (x_1)=1+\lambda$.

Consider the oriented pro-p group $(U,\theta \vert _U)$, and set $K=K_{\theta \vert _U}(U)$, ${\bar{U}}=U/K$. Our goal is to show that the oriented pro-p group $({\bar{U}},(\theta \vert _U)_{/K})$ is not $(\theta \vert _U)_{/K}$-abelian, so that $(U,\theta \vert _U)$ is not Kummerian by Proposition 2.8, and thus $(G,\theta )$ is not 1-cyclotomic.

Since $K\subseteq \Phi (U)$, by Proposition 7.3 the set $\mathcal {Y}_{{\bar{U}}}=\{wK\,\mid \,w\in \mathcal {Y}_U\}$ is a minimal generating set of ${\bar{U}}$. Now, since $\theta (t)=\theta (x_1)=(1+\lambda )^{-1}$, one has $w^t\equiv w^{1+\lambda }\bmod K$ for every $w\in U$. Therefore, from (7.2), and from the fact that $[x_4,x_5]\in {{\,\textrm{Ker}\,}}(\theta \vert _U)'\subseteq K$, one obtains

$$\begin{aligned}{}[x_1,x_2]^{-1}\left( [x_1,x_2]^{x_1}\right) ^t\equiv [x_1,x_2]^{-1}\left( [x_1,x_2]^{x_1}\right) ^{(1+\lambda )^{-1}}\equiv 1\mod K, \end{aligned}$$

and consequently

$$\begin{aligned} \begin{aligned}{}[x_1,x_2]^{x_1}&\equiv [x_1,x_2]^{1+\lambda }\mod K,\\ [x_1,x_2]^{x_1^2}&\equiv [x_1,x_2]^{(1+\lambda )^2}\mod K,\\&\quad \vdots \\ [x_1,x_2]^{x_1^{p-1}}&\equiv [x_1,x_2]^{(1+\lambda )^{p-1}}. \end{aligned} \end{aligned}$$

(7.5)

Set

$$\begin{aligned}\mu =(1+\lambda )^0+(1+\lambda )^1+\cdots +(1+\lambda )^{p-1}=\dfrac{(1+\lambda )^p-1}{\lambda }.\end{aligned}$$

Then $\mu \ne 0$ (as $\lambda \ne 0$), and $p\mid \mu$. Since $[x_1,x_2]=(x_2^{x_1})^{-1}x_2$, replacing the coset $x_2^{x_1}K$ with the coset $[x_1,x_2]K$ in $\mathcal {Y}_{{\bar{U}}}$ yields another minimal generating set—let us call it $\mathcal {Y}'_{{\bar{U}}}$—of ${\bar{U}}$. Now, from (7.5) one obtains

$$\begin{aligned} \begin{aligned}{}[u,x_2]&=[x_1,x_2]^{x_1^{p-1}}\cdots [x_1,x_2]^{x_1}\cdot [x_1,x_2]\\&\equiv [x_1,x_2]^{(1+\lambda )^{p-1}}\cdots [x_1,x_2]^{1+\lambda }\cdot [x_1,x_2]\mod K\\&\equiv [x_1,x_2]^{\mu }\mod K \end{aligned} \end{aligned}$$

—observe that $[x_1,x_2]^{x_i^h}\in {{\,\textrm{Ker}\,}}(\theta \vert _U)$ for every h, and thus all such elements commute modulo K. Therefore, one has the relation

$$\begin{aligned} \left( [x_1,x_2]K\right) ^\mu =[uK,x_2K] \end{aligned}$$

between elements of the minimal generating set $\mathcal {Y}'_{{\bar{U}}}$, and by [11, Theorem 8.1] this relation prevents the oriented pro-p group $({\bar{U}},(\theta \vert _U)_{/K})$ from being Kummerian—and thus also $(\theta \vert _U)_{/K}$-abelian. $\square$

From Theorem 1.4, we obtain a new family of pro-p groups which cannot complete into 1-cyclotomic oriented pro-p groups.

Corollary 7.4

Let G be the pro-p group with presentation

$$\begin{aligned} G=\left\langle x_1,\ldots ,x_n,x_{n+1},x_{n+2} \,\mid \, \left[ \left[ \ldots [[x_1,x_2],x_3],\ldots x_{n-1}\right] ,x_n\right] [x_{n+1},x_{n+2}]=1\right\rangle , \end{aligned}$$

with $n\ge 3$. Then G cannot complete into a 1-cyclotomic oriented pro-p group with torsion-free orientation.

Proof

Set $y=[\ldots [x_1,x_2],\ldots x_{n-2}]$, and let H be the subgroup of G generated by $\{y,x_{n-1},\ldots ,x_{n+2}\}$. Then

$$\begin{aligned} H=\langle \,y,x_{n-1},\ldots ,x_{n+2}\,\mid \,[[y,x_{n-1}],x_n][x_{n+1},x_{n+2}]\,\rangle \end{aligned}$$

is isomorphic to the Minač–Tân pro-p group, and hence it cannot complete into a 1-cyclotomic oriented pro-p group with torsion-free orientation by Theorem 1.4. $\square$

The following question remains open (cf. [2, Example 3.2]).

Question 7.5

Is the Minač–Tân pro-p group G a Bloch–Kato pro-p group? Namely, is the $\mathbb {Z}/p\mathbb {Z}$-cohomology algebra of every closed subgroup of G a quadratic algebra?