On discrete homology of a free pro-$p$-group

TL;DR

This paper proves that the second discrete homology group of a finitely generated free pro-$p$-group is uncountably infinite-dimensional over $Z/p$, resolving a question posed by Bousfield.

Contribution

It establishes the uncountability of the second discrete homology group for free pro-$p$-groups, providing a significant insight into their homological properties.

Findings

$H_2(hat_p,Z/p)$ is uncountable

Answers Bousfield's problem

Advances understanding of pro-$p$-group homology

Abstract

For a prime $p$, let $\hat F_p$ be a finitely generated free pro-$p$-group of rank $\geq 2$. We show that the second discrete homology group $H_2(\hat F_p,\mathbb Z/p)$ is an uncountable $\mathbb Z/p$-vector space. This answers a problem of A.K. Bousfield.

Equations

$$H_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\to H_2^{\mathsf{cont}}(G,\mathbb{Z}/p),$$

$$H_2^{\mathsf{cont}}({\hat{F}}_p,\mathbb{Z}/p)=0.$$

$$H_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H_2^{\mathsf{disc}}({\hat{F}}_p,\mathbb{Z}/p)\stackrel{{\pi }_{\ast }}{⟶}{H}_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\stackrel{\phi }{⟶}{H}_2^{\mathsf{cont}}(G,\mathbb{Z}/p)⟶0.$$

$$\mathbb{Z}/p[[x]]{\otimes }_{\mathbb{Z}/p[C\otimes {\mathbb{Z}}_p]}\mathbb{Z}/p[[x]]⟶\mathbb{Z}/p[[x]]$$

$$(\mathbb{Z}/p{)}^2\wr \mathbb{Z}=⟨a,b,c\mid [b,b^{a^i}]=[c,c^{a^i}]=[b,c^{a^i}]=b^p=c^p=1,i\in \mathbb{Z}⟩,$$

$$\mathbb{Z}/p[[x]]{\otimes }_{\mathbb{Z}/p[C\otimes {\mathbb{Z}}_p]}\mathbb{Z}/p[[x]]\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H_2(G,\mathbb{Z}/p)⟶H_2(G/H,\mathbb{Z}/p)⟶\frac{H\cap ([G,G]G^p)}{[H,G]H^p}⟶0.$$

$$H_2(G,\mathbb{Z}/p)⟶H_2(G/H,\mathbb{Z}/p)⟶H_1(H,\mathbb{Z}/p{)}_G⟶H_1(G,\mathbb{Z}/p)$$

$$H_2^{\mathsf{cont}}(G,\mathbb{Z}/p)⟶H_2^{\mathsf{cont}}(G/H,\mathbb{Z}/p)⟶\frac{H\cap \overline{([G,G]G^p)}}{\overline{([H,G]H^p)}}⟶0.$$

$$H_2^{\mathsf{cont}}(G)⟶H_2^{\mathsf{cont}}(G)⟶H_0^{\mathsf{cont}}(G,H_1^{\mathsf{cont}}(H))⟶H_1^{\mathsf{cont}}(G).$$

$$H_{\mathsf{cont}}^1(G)=Hom(G/\overline{[G,G]G^p},\mathbb{Z}/p)=Hom(G/\overline{[G,G]G^p},\mathbb{Q}/\mathbb{Z}),$$

$$\phi :H_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\to H_2^{\mathsf{cont}}(G,\mathbb{Z}/p).$$

$$Q^{\mathsf{disc}}:=\mathsf{Coker}(H_2(G,\mathbb{Z}/p)\to H_2(G/H,\mathbb{Z}/p))$$

$$Q^{\mathsf{cont}}:=\mathsf{Coker}(H_2^{\mathsf{cont}}(G,\mathbb{Z}/p)\to H_2^{\mathsf{cont}}(G/H,\mathbb{Z}/p)).$$

$$H_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H_2^{\mathsf{disc}}({\hat{F}}_p,\mathbb{Z}/p)\stackrel{{\pi }_{\ast }}{⟶}{H}_2^{\mathsf{disc}}(G,\mathbb{Z}/p)\stackrel{\phi }{⟶}{H}_2^{\mathsf{cont}}(G,\mathbb{Z}/p)⟶0$$

$$\underset{i\to \infty }{\lim }|{\mathsf{A}}^i|/p^{p^i}=0.$$

$$\tau :C\otimes {\mathbb{Z}}_p\to \mathbb{Z}/p[[x]]$$

$$\frac{{\alpha }_1{a}_1+\cdots +{\alpha }_n{a}_n}{{\beta }_1{b}_1+\cdots +{\beta }_n{b}_n},$$

$$\bar{r}\cdot ({\beta }_1{\bar{b}}_1+\cdots +{\beta }_n{\bar{b}}_n)={\alpha }_1{\bar{a}}_1+\dots {\alpha }_n{\bar{a}}_n.$$

$$|{\mathsf{A}}_{\alpha ,\beta ,k}^i|\le p^{2in+p^k}.$$

$${\mathsf{A}}_{\alpha ,\beta ,k,l}={\mathsf{A}}_{\alpha ,\beta ,k}\cdot v_1+\cdots +{\mathsf{A}}_{\alpha ,\beta ,k}\cdot v_l.$$

$$\underset{i\to \infty }{\lim }|{\mathsf{A}}_{\alpha ,\beta ,k,l}^i|/p^{p^i}=0.$$

$$\mathbb{Z}/p[[x]]{\otimes }_{\mathbb{Z}/p[C\otimes {\mathbb{Z}}_p]}\mathbb{Z}/p[[x]]⟶\mathbb{Z}/p[[x]]$$

$$\mu :\mathbb{Z}/p[[x]]{\otimes }_{R}K\to \mathbb{Z}/p((x)).$$

$$\kappa :\mathbb{Z}/p((x))\to \mathbb{Z}/p[[x]]{\otimes }_{R}K$$

$$\kappa (\sum _{i=-n}^{\infty }{\alpha }_i{x}^i)=\sum _{i=0}^{\infty }{\alpha }_{i+n}{x}^i\otimes x^{-n}.$$

$$\mathbb{Z}/p[[x]]{\otimes }_{R}K\cong \mathbb{Z}/p((x)).$$

$$\mathbb{Z}/p[[x]]{\otimes }_R\mathbb{Z}/p[[x]]⟶\mathbb{Z}/p[[x]]$$

$$\mathbb{Z}/p((x)){\otimes }_K\mathbb{Z}/p((x))⟶\mathbb{Z}/p((x)).$$

Full text

1. Introduction

Let $p$ be a prime. For a profinite group $G$, there is a natural comparison map

[TABLE]

which connects discrete and continuous homology groups of $G$. Here $H_{2}^{\sf disc}(G,{\mathbb{Z}}/p)=H_{2}(G,{\mathbb{Z}}/p)$ is the second homology group of $G$ with ${\mathbb{Z}}/p$-coefficients, where $G$ is viewed as a discrete group. The continuous homology $H_{2}^{\sf cont}(G,{\mathbb{Z}}/p)$ can be defined as the inverse limit $\varprojlim H_{2}(G/U,{\mathbb{Z}}/p),$ where $U$ runs over all open normal subgroups of $G$. The above comparison map $H_{2}^{\sf disc}\to H_{2}^{\sf cont}$ is the inverse limit of the coinflation maps $H_{2}(G,{\mathbb{Z}}/p)\to H_{2}(G/U,{\mathbb{Z}}/p)$ (see Theorem 2.1 [6]).

The study of the comparison map for different types of pro-$p$-groups is a fundamental problem in the theory of profinite groups (see [6] for discussion and references). It is well known that for a finitely generated free pro-$p$-group $\hat{F}_{p}$,

[TABLE]

A.K. Bousfield posed the following question in [2], Problem 4.11 (case $R=\mathbb{Z}/n$):

Problem. (Bousfield) Does $H_{2}^{\sf disc}(\hat{F}_{n},\mathbb{Z}/n)$ vanish when $F$ is a finitely generated free group?

Here $\hat{F}_{n}$ is the ${\mathbb{Z}}/n$-completion of $F$, which is isomorphic to the product of pro-$p$-completions $\hat{F}_{p}$ over prime factors of $n$ (see Prop. 12.3 [2]). That is, the above problem is completely reduced to the case of homology groups $H_{2}^{\sf disc}(\hat{F}_{p},\mathbb{Z}/p)$ for primes $p$ and since $H_{2}^{\sf cont}(\hat{F}_{p},\mathbb{Z}/p)=0$, the problem becomes a question about non-triviality of the kernel of the comparison map for $\hat{F}_{p}$.

In [3], A.K. Bousfield proved that, for a finitely generated free pro-$p$-group $\hat{F}_{p}$ on at least two generators, the group $H_{i}^{\sf disc}(\hat{F}_{p},\mathbb{Z}/p)$ is uncountable for $i=2$ or $i=3$, or both. In particular, the wedge of two circles $S^{1}\vee S^{1}$ is a $\mathbb{Z}/p$-bad space in the sense of Bousfield-Kan.

The group $H_{2}^{\sf disc}(\hat{F}_{p},{\mathbb{Z}}/p)$ plays a central role in the theory of $H{\mathbb{Z}}/p$-localizations developed in [2]. It follows immediately from the definition of $H{\mathbb{Z}}/p$-localization that, for a free group $F$, $H_{2}^{\sf disc}(\hat{F}_{p},{\mathbb{Z}}/p)=0$ if and only if $\hat{F}_{p}$ coincides with the $H{\mathbb{Z}}/p$-localization of $F$.

In this paper we answer Bousfield’s problem over ${\mathbb{Z}}/p$. Our main result is the following

Main Theorem. For a finitely generated free pro-$p$-group $\hat{F}_{p}$ of rank $\geq 2$, $H_{2}^{\sf disc}(\hat{F}_{p},\mathbb{Z}/p)$ is uncountable.

There are two cases in the problem of Bousfield, $R={\mathbb{Z}}/n$ and $R={\mathbb{Q}}.$ We give the answer for the case of $R={\mathbb{Z}}/n$. The problem over ${\mathbb{Q}}$ is still open: we do not know whether $H_{2}^{\sf disc}(\hat{F}_{\mathbb{Q}},{\mathbb{Q}})$ vanishes or not.

The proof is organized as follows.

In Section 2 we consider properties of discrete and continuous homology of profinite groups. Using a result of Nikolov and Segal [10, Th 1.4], we show that for a finitely generated profinite group $G$ and a closed normal subgroup $H$ the cokernels of the maps $H_{2}^{\sf disc}(G,{\mathbb{Z}}/p)\to H_{2}^{\sf disc}(G/H,{\mathbb{Z}}/p)$ and $H_{2}^{\sf cont}(G,{\mathbb{Z}}/p)\to H_{2}^{\sf cont}(G/H,{\mathbb{Z}}/p)$ coincide (Theorem 2.5):

[TABLE]

As a corollary we obtain (Corollary 2.6) that, for a finitely generated free pro-$p$-group $\hat{F}_{p}$, a continuous epimorphism $\pi:\hat{F}_{p}\twoheadrightarrow G$ to a pro-$p$-group induces the exact sequence

[TABLE]

That is, for proving that, for a free group $F$, $H_{2}(\hat{F}_{p},{\mathbb{Z}}/p)\neq 0$, it is enough to find a discrete epimorphism $F\twoheadrightarrow G$ such that, the comparison map of the second homology groups of the pro-$p$-completion of $G$ has a nonzero kernel. Observe that, the statements in Section 2 significantly use the theory of profinite groups and there is no direct way to generalize them for pronilpotent groups. In particular, we don’t see how to prove that, $H_{2}(\hat{F}_{{\mathbb{Z}}},\mathbb{Z}/p)\neq 0$, where $\hat{F}_{{\mathbb{Z}}}$ is the pronilpotent completion of $F$.

Section 3 follows the ideas of Bousfield from [3]. Consider the ring of formal power series ${\mathbb{Z}}/p[\![x]\!]$, and the infinite cyclic group $C:=\langle t\rangle$. We will use the multiplicative notation of the $p$-adic integers $C\otimes\mathbb{Z}_{p}=\{t^{\alpha},\ \alpha\in\mathbb{Z}_{p}\}$. Consider the continuous multiplicative homomorphism $\tau:C\otimes\mathbb{Z}_{p}\to{\mathbb{Z}}/p[\![x]\!]$ sending $t$ to $1-x$. The main result of Section 3 is Proposition 3.3, which claims that the kernel of the multiplication map

[TABLE]

is uncountable.

Our main example is based on the $p$-lamplighter group $\mathbb{Z}/p\wr\mathbb{Z}$, a finitely generated but not finitely presented group, which plays a central role in the theory of metabelian groups. The homological properties of the $p$-lamplighter group are considered in [9]. The profinite completion of the $p$-lamplighter group is considered in [7], it is shown there that it is a semi-similar group generated by finite automaton. We consider the double lamplighter group:

[TABLE]

Denote by ${\mathcal{D}\mathcal{L}}$ the pro-$p$-completion of the double lamplighter group. It follows from direct computations of homology groups, that there is a diagram (in the above notations)

[TABLE]

where the left vertical arrow is a split monomorphism and the upper horisontal map is the multiplication map (see proof of Theorem 4.3). This implies that, for the group ${\mathcal{D}\mathcal{L}}$, the comparison map $H_{2}^{\sf disc}({\mathcal{D}\mathcal{L}},{\mathbb{Z}}/p)\to H_{2}^{\sf cont}({\mathcal{D}\mathcal{L}},{\mathbb{Z}}/p)$ has an uncountable kernel. Since the double lamplighter group is 3-generated, the sequence (1.1) implies that, for a free group $F$ with at least three generators, $H_{2}(\hat{F}_{p},\mathbb{Z}/p)$ is uncountable. Finally, we use Lemma 11.2 of [3] to get the same result for a two generated free group $F$.

In [2], A.K. Bousfield formulated the following generalization of the above problem for the class of finitely presented groups (see Problem 4.10 [2], the case $R=\mathbb{Z}/n$). Let $G$ be a finitely presented group, is it true that $H\mathbb{Z}/p$-localization of $G$ equals to its pro-$p$-completion $\hat{G}_{p}$? [The Problem is formulated for $H\mathbb{Z}/n$-localization, but it is reduced to the case of a prime $n=p$.] It follows immediately from the definition of $H\mathbb{Z}/p$-localization that, this problem can be reformulated as follows: is it true that, for a finitely presented group $G$, the natural homomorphism $H_{2}(G,\mathbb{Z}/p)\to H_{2}(\hat{G}_{p},\mathbb{Z}/p).$ It is shown in [2] that this is true for the class of poly-cyclic groups. The same is true for finitely presented metabelian groups [8]. Main Theorem of the present paper implies that, for any finitely presented group $P$, which maps epimorphically onto the double lamplighter group, the natural map $H_{2}(P,\mathbb{Z}/p)\to H_{2}(\hat{P}_{p},\mathbb{Z}/p)$ has an uncountable cokernel.

2. Discrete and continuous homology of profinite groups

For a profinite group $G$ and a normal subgroup $H$, denote by $\overline{H}$ the closure of $H$ in $G$ in profinite topology.

Theorem 2.1** ([10, Th 1.4]).**

Let $G$ be a finitely generated profinite group and $H$ be a closed normal subgroup of $G.$ Then the subgroup $[H,G]$ is closed in $G.$

Corollary 2.2**.**

Let $G$ be a finitely generated profinite group and $H$ be a closed normal subgroup of $G.$ Then the subgroup $[H,G]\cdot H^{p}$ is closed in $G.$

Proof.

Consider the abelian profinite group $H/[H,G].$ Then the $p$-power map $H/[H,G]\to H/[H,G]$ is continuous and its image is equal to $([H,G]\cdot H^{p})/[H,G].$ Hence $([H,G]\cdot H^{p})/[H,G]$ is a closed subgroup of $H/[H,G].$ Using that the preimage of a closed set under continuous function is closed we obtain that $[H,G]\cdot H^{p}$ is closed. ∎

Lemma 2.3** (mod-$p$ Hopf’s formula).**

Let $G$ be a (discrete) group and $H$ be its normal subgroup. Then there is a natural exact sequence

[TABLE]

Proof.

It follows from the five term exact sequence

[TABLE]

and the equations $H_{1}(H,{\mathbb{Z}}/p)_{G}=H/([H,G]H^{p})$ and $H_{1}(G,{\mathbb{Z}}/p)=G/([G,G]G^{p}).$ ∎

Lemma 2.4** (profinite mod-$p$ Hopf’s formula).**

Let $G$ be a profinite group and $H$ be its closed normal subgroup. Then there is a natural exact sequence

[TABLE]

Proof.

For the sake of simplicity we set $H_{*}(-)=H_{*}^{{\sf discr}}(-,{\mathbb{Z}}/p)$ and $H_{*}^{\sf cont}(-):=H^{\sf cont}_{*}(-,{\mathbb{Z}}/p).$ Consider the five term exact sequence (Corollary 7.2.6 of [11])

[TABLE]

Continuous homology and cohomology of profinite groups are Pontryagin dual to each other (Proposition 6.3.6 of [11]). There are isomorphisms

[TABLE]

where ${\it Hom}$ denotes the set of continuous homomorphisms (see [12, I.2.3]). It follows that $H_{1}^{\sf cont}(G)=G/\overline{[G,G]G^{p}}.$ Similarly $H_{1}^{\sf cont}(H)=H/\overline{[H,H]H^{p}}.$ Lemma 6.3.3. of [11] implies that $H_{0}^{\sf cont}(G,M)=M/\overline{\langle m-mg\mid m\in M,g\in G\rangle}$ for any profinite $({\mathbb{Z}}/p[G])^{\wedge}$-module $M.$ Therefore $H^{\sf cont}_{0}(G,H^{\sf cont}_{1}(H))=H/\overline{[H,H]H^{p}}.$ The assertion follows. ∎

We denote by $\varphi$ the comparison map

[TABLE]

Theorem 2.5**.**

Let $G$ be a finitely generated profinite group and $H$ a closed normal subgroup of $G$. Denote

[TABLE]

Then the comparison maps $\varphi$ induce an isomorphism $Q^{\sf disc}\cong Q^{\sf cont}:$

[TABLE]

Proof.

This follows from Lemma 2.3, Lemma 2.4 and Corollary 2.2. ∎

Corollary 2.6**.**

Let $G$ be a finitely generated pro-$p$-group and $\pi:\hat{F}_{p}\twoheadrightarrow G$ be a continuous epimorphism from the pro-$p$-completion of a finitely generated free group $F.$ Then the sequence

[TABLE]

is exact.

Proof.

This follows from Theorem 2.5 and the fact that $H_{2}^{\sf cont}(\hat{F}_{p},{\mathbb{Z}}/p)=0.$ ∎

3. Technical results about the ring of power series ${\mathbb{Z}}/p[\![x]\!]$

In this section we follow to ideas of Bousfield written in Lemmas 10.6, 10.7 of [3]. The goal of this section is to prove Proposition 3.3

We use the following notation: $C=\langle t\rangle$ is the infinite cyclic group; $C\otimes{\mathbb{Z}}_{p}$ is the group of $p$-adic integers written multiplicatively as powers of the generator $C\otimes{\mathbb{Z}}_{p}=\{t^{\alpha}\mid\alpha\in{\mathbb{Z}}_{p}\};$ ${\mathbb{Z}}/p[\![x]\!]$ is the ring of power series; ${\mathbb{Z}}/p(\!(x)\!)$ is the field of formal Laurent series.

Lemma 3.1**.**

Let ${\sf A}$ be a subset of ${\mathbb{Z}}/p[\![x]\!].$ Denote by ${\sf A}^{i}$ the image of ${\sf A}$ in ${\mathbb{Z}}/p[x]/(x^{p^{i}}).$ Assume that

[TABLE]

Then the interior of ${\mathbb{Z}}/p[\![x]\!]\setminus{\sf A}$ is dense in ${\mathbb{Z}}/p[\![x]\!].$

Proof.

Take any power series $f$ and any its neighbourhood of the form $f+(x^{p^{s}}).$ Then for any $i$ the open set $f+(x^{p^{s}})$ is the disjoint union of smaller open sets $\bigcup_{t=1}^{p^{p^{i}}}f+f_{t}+(x^{p^{s+i}}),$ where $f_{t}$ runs over representatives of $(x^{p^{s}})/(x^{p^{s+i}}).$ Chose $i$ so that $|{\sf A}^{s+i}|/p^{p^{i+s}}\leq p^{-p^{s}}.$ Then $|{\sf A}^{s+i}|\leq p^{p^{i}}.$ Hence the number of elements in ${\sf A}^{i+s}$ is lesser the number of open sets $f+f_{t}+(x^{p^{s+i}})$. It follows that there exists $t$ such that ${\sf A}\cap(f+f_{t}+(x^{p^{s+i}}))=\emptyset.$ The assertion follows. ∎

Denote by

[TABLE]

the continuous multiplicative homomorphism sending $t$ to $1-x.$ It is well defined because $(1-x)^{p^{i}}=1-x^{p^{i}}.$

Lemma 3.2**.**

Let $K$ be the subfield of ${\mathbb{Z}}/p(\!(x)\!)$ generated by the image of $\tau$. Then the degree of the extension $[{\mathbb{Z}}/p(\!(x)\!):K]$ is uncountable.

Proof.

Denote the image of the map $\tau:C\otimes{\mathbb{Z}}_{p}\to{\mathbb{Z}}/p[\![x]\!]$ by ${\sf A}.$ Let $\alpha=(\alpha_{1},\dots,\alpha_{n})$ and $\beta=(\beta_{1},\dots,\beta_{n})$, where $\alpha_{1},\dots,\alpha_{n},\beta_{1},\dots,\beta_{n}\in{\mathbb{Z}}/p,$ and $k\geq 1.$ Denote by ${\sf A}_{\alpha,\beta,k}$ the subset of ${\mathbb{Z}}/p[\![x]\!]$ consisting of elements that can be written in the form

[TABLE]

where $a_{1},\dots,a_{n},b_{1},\dots,b_{n}\in{\sf A}$ and $\beta_{1}b_{1}+\dots+\beta_{n}b_{n}\notin(x^{p^{k}}).$ Then $K\cap{\mathbb{Z}}/p[\![x]\!]=\bigcup_{\alpha,\beta,k}{\sf A}_{\alpha,\beta,k}.$

Fix some $\alpha,\beta,k.$ Take $i\geq k$ and consider the images of ${\sf A}$ and ${\sf A}_{\alpha,\beta,k}$ in ${\mathbb{Z}}/p[x]/(x^{p^{i}}).$ Denote them by ${\sf A}^{i}$ and ${\sf A}_{\alpha,\beta,k}^{i}.$ Obviously ${\sf A}^{i}$ is the image of the map $C/C^{p^{i}}\to{\mathbb{Z}}/p[x]/(x^{p^{i}})$ that sends $t$ to $1-x.$ Then ${\sf A}^{i}$ consists of $p^{i}$ elements. Fix some elements $\bar{a}_{1},\dots,\bar{a}_{n},\bar{b}_{1},\dots,\bar{b}_{n}\in{\sf A}^{i}$ that have preimages $a_{1},\dots,a_{n},b_{1},\dots,b_{n}\in{\sf A}$ such that the ratio (3.1) is in ${\sf A}_{\alpha,\beta,k}.$ For any such preimages $a_{1},\dots,a_{n},b_{1},\dots,b_{n}\in{\sf A}$ the image $\bar{r}$ of the ratio (3.1) satisfies the equation

[TABLE]

Since $\beta_{1}\bar{b}_{1}+\dots+\beta_{n}\bar{b}_{n}\notin(x^{p^{k}}),$ the annihilator of $\beta_{1}\bar{b}_{1}+\dots+\beta_{n}\bar{b}_{n}$ consist of no more than $p^{p^{k}}$ elements and the equation has no more than $p^{p^{k}}$ solutions. Then we have no more than $p^{2in}$ variants of collections $\bar{a}_{1},\dots,\bar{a}_{n},\bar{b}_{1},\dots,\bar{b}_{n}\in{\sf A}^{i}$ and for any such variant there are no more than $p^{p^{k}}$ variants for the image of the ratio. Therefore

[TABLE]

Take any sequence of elements $v_{1},v_{2},\dots\in{\mathbb{Z}}/p(\!(x)\!)$ and prove that $\sum_{m=1}^{\infty}Kv_{m}\neq{\mathbb{Z}}/p(\!(x)\!).$ Note that $x\in K$ because $t\mapsto 1-x.$ Multiplying the elements $v_{1},v_{2},v_{3},\dots$ by powers of $x,$ we can assume that $v_{1},v_{2},\dots\in{\mathbb{Z}}/p[\![x]\!].$ Fix some $\alpha,\beta,k$ as above. Set

[TABLE]

Then $\sum_{m=1}^{\infty}Kv_{m}=\bigcup_{\alpha,\beta,k,l,j}{\sf A}_{\alpha,\beta,k,l}\cdot x^{-j}.$ Denote by ${\sf A}_{\alpha,\beta,k,l}^{i}$ the image of ${\sf A}_{\alpha,\beta,k,l}$ in ${\mathbb{Z}}/p[x]/(x^{i}).$ Then (3.2) implies $|{\sf A}_{\alpha,\beta,k,l}^{i}|\leq p^{(2in+p^{k})l}.$ Therefore

[TABLE]

By Lemma 3.1 the interior of the complement of ${\sf A}_{\alpha,\beta,k,l}$ is dense in ${\mathbb{Z}}/p[\![x]\!]$. By the Baire theorem $\sum_{m=1}^{\infty}Kv_{m}=\bigcup_{\alpha,\beta,k,l,j}{\sf A}_{\alpha,\beta,k,l}\cdot x^{-j}$ has empty interior. In particular $\sum_{m=1}^{\infty}Kv_{m}\neq{\mathbb{Z}}/p(\!(x)\!).$ ∎

Proposition 3.3**.**

Consider the ring homomorphism ${\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}]\to{\mathbb{Z}}/p[\![x]\!]$ induced by $\tau$. Then the kernel of the multiplication map

[TABLE]

is uncountable.

Proof.

As in Lemma 3.2 we denote by $K$ the subfield of ${\mathbb{Z}}/p(\!(x)\!)$ generated by the image of $C\otimes{\mathbb{Z}}_{p}.$ Since $t\mapsto 1-x,$ we have $x,x^{-1}\in K.$ Set $R:=K\cap{\mathbb{Z}}/p[\![x]\!].$ Note that the image of ${\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}]$ lies in $R.$ Consider the multiplication map

[TABLE]

We claim that this is an isomorphism. Construct the map in the inverse direction

[TABLE]

given by

[TABLE]

Since we have $ax\otimes b=a\otimes xb,$ $\kappa$ does not depend on the choice of $n$, we just have to chose it big enough. Using this we get that $\kappa$ is well defined. Obviously $\mu\kappa={\sf id}.$ Chose $a\otimes b\in{\mathbb{Z}}/p[\![x]\!]\otimes_{R}K.$ Then $b=b_{1}b_{2}^{-1},$ where $b_{1},b_{2}\in R.$ Since $b_{2}$ is a power series, we can chose $n$ such that $b_{2}=x^{n}b_{3},$ where $b_{3}$ is a power series with nontrivial constant term. Then $b_{3}$ is invertible in the ring of power series and $b_{3},b_{3}^{-1}\in R$ because $x\in K.$ Hence $a\otimes b=ab_{3}^{-1}b_{3}\otimes b=ab_{3}^{-1}\otimes x^{-n}b_{1}=ab_{1}b_{3}^{-1}\otimes x^{-n}.$ Using this presentation, we see that $\kappa\mu={\sf id}.$ Therefore

[TABLE]

Since the image of ${\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}]$ lies in $R,$ the tensor product ${\mathbb{Z}}/p[\![x]\!]\otimes_{R}{\mathbb{Z}}/p[\![x]\!]$ is a quotient of the tensor product ${\mathbb{Z}}/p[\![x]\!]\otimes_{{\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}]}{\mathbb{Z}}/p[\![x]\!]$ and it is enough to prove that the kernel of

[TABLE]

is uncountable.

For any ring homomorphism $R\to S$ and any $R$-modules $M,N$ there is an isomorphism $(M\otimes_{R}N)\otimes_{R}S=(M\otimes_{R}S)\otimes_{S}(N\otimes_{R}S).$ Using this and the isomorphism (3.4), we obtain that after application of $-\otimes_{R}K$ to (3.5) we have

[TABLE]

Assume the contrary that the kernel of the map (3.5) is countable (countable = countable or finite). It follows that the linear map (3.6) has countably dimensional kernel. Finally, note that the homomorphism

[TABLE]

given by $a\wedge b\mapsto a\otimes b-b\otimes a$ is a monomorphism, its image lies in the kernel and the dimension of $\Lambda^{2}_{K}{\mathbb{Z}}/p(\!(x)\!)$ over $K$ is uncountable because $[{\mathbb{Z}}/p(\!(x)\!):K]$ is uncountable (Lemma 3.2). A contradiction follows. ∎

4. Double lamplighter pro-$p$-group

Let $A$ be a finitely generated free abelian group written multiplicatively; ${\mathbb{Z}}/p[A]$ be its group algebra; $I$ be its augmentation ideal and $M$ be a ${\mathbb{Z}}/p[A]$-module. Then we denote by $\hat{M}=\varprojlim M/MI^{i}$ its $I$-adic completion. We embed $A$ into the pro-$p$-group $A\otimes{\mathbb{Z}}_{p}.$ We use the following ’multiplicative’ notation $a^{\alpha}:=a\otimes\alpha$ for $a\in A$ and $\alpha\in{\mathbb{Z}}_{p}.$ Note that for any $a\in A$ the power $a^{p^{i}}$ acts trivially on $M/MI^{p^{i}}$ because $1-a^{p^{i}}=(1-a)^{p^{i}}\in I^{p^{i}}.$ Then we can extend the action of $A$ on $\hat{M}$ to the action of $A\otimes{\mathbb{Z}}_{p}$ on $\hat{M}$ in a continuous way.

The proof of the following lemma can be found in [8] but we add it here for completeness.

Lemma 4.1**.**

Let $A$ be a finitely generated free abelian group and $M$ be a finitely generated ${\mathbb{Z}}/p[A]$-module. Then

[TABLE]

Proof.

The first isomorphism is proven in [4]. Since ${\mathbb{Z}}/p[A]$ in Noetheran, it follows that $H_{n}(A,\hat{M})$ is a finite ${\mathbb{Z}}/p$-vector space for any $n$. Prove the second isomorphism. The action of $A\otimes{\mathbb{Z}}_{p}$ on $\hat{M}$ gives an action of $A\otimes{\mathbb{Z}}_{p}$ on $H_{*}(A,\hat{M})$ such that $A$ acts trivially on $H_{*}(A,\hat{M}).$ Then we have a homomorphism from $A\otimes{\mathbb{Z}}/p$ to a finite group of automorphisms of $H_{n}(A,\hat{M}),$ whose kernel contains $A.$ Since any subgroup of finite index in $A\otimes{\mathbb{Z}}_{p}$ is open (see Theorem 4.2.2 of [11]) and $A$ is dense in $A\otimes{\mathbb{Z}}_{p},$ we obtain that the action of $A\otimes{\mathbb{Z}}_{p}$ on $H_{*}(C,\hat{M})$ is trivial. Note that ${\mathbb{Z}}_{p}/{\mathbb{Z}}$ is a divisible torsion free abelian group, and hence $A\otimes({\mathbb{Z}}_{p}/{\mathbb{Z}})\cong{\mathbb{Q}}^{\oplus{\bf c}},$ where ${\bf c}$ is the continuous cardinal. Then the second page of the spectral sequence of the short exact sequence $A\rightarrowtail A\otimes{\mathbb{Z}}_{p}\twoheadrightarrow{\mathbb{Q}}^{\oplus{\bf c}}$ with coefficients in $\hat{M}$ is $H_{n}({\mathbb{Q}}^{\oplus{\bf c}},H_{m}(A,\hat{M})),$ where $L_{m}:=H_{m}(A,\hat{M})$ is a trivial ${\mathbb{Z}}/p[{\mathbb{Q}}^{\oplus{\bf c}}]$-module. Then by universal coefficient theorem we have

[TABLE]

Since $\Lambda^{n}({\mathbb{Q}}^{\oplus{\bf c}})$ is torsion free and $L_{m}$ is a ${\mathbb{Z}}/p$-vector space, we get $\Lambda^{n}({\mathbb{Q}}^{\oplus{\bf c}})\otimes L_{m}=0$ and ${\sf Tor}(\Lambda^{n-1}({\mathbb{Q}}^{\oplus{\bf c}}),L_{m})=0.$ It follows that $H_{n}({\mathbb{Q}}^{\oplus{\bf c}},L_{m})=0$ for $n\geq 1$ and $H_{0}({\mathbb{Q}}^{\oplus{\bf c}},L_{m})=L_{m}.$ Then the spectral sequence consists only on of one column, and hence $H_{*}(A\otimes{\mathbb{Z}}_{p},\hat{M})=H_{*}(A,\hat{M}).$ ∎

Lemma 4.2**.**

Let $A$ be an abelian group, $M$ be a ${\mathbb{Z}}[A]$-module and $\sigma_{M}:M\to M$ be an automorphism of the underlying abelian group such that $\sigma_{M}(ma)=\sigma_{M}(m)a^{-1}$ for any $m\in M$ and $a\in A.$ Then there is an isomorphism

[TABLE]

given by

[TABLE]

Proof.

Consider the isomorphism $\Phi:M\otimes M\to M\otimes M$ given by $\Phi(m\otimes m^{\prime})=m\otimes\sigma(m^{\prime}).$ The group of coinvariants $(M\otimes M)_{A}$ is the quotient of $M\otimes M$ by the subgroup $R$ generated by elements $ma\otimes m^{\prime}a-m\otimes m^{\prime},$ where $a\in A$ and $m,m^{\prime}\in M.$ We can write the generators of $R$ in the following form: $ma\otimes m^{\prime}$ $-$ $m\otimes m^{\prime}a^{-1}.$ Then $\Phi(R)$ is generated by $ma\otimes\sigma_{M}(m^{\prime})$ $-$ $m\otimes\sigma_{M}(m^{\prime})a.$ Using that $\sigma_{M}$ is an automorphism, we can rewrite the generators of $R$ as follows: $ma\otimes m^{\prime}-m\otimes m^{\prime}a.$ Taking linear combinations of the generators of $\Phi(R)$ we obtain that $\Phi(R)$ is generated by elements $m\lambda\otimes m^{\prime}-m\otimes m^{\prime}\lambda,$ where $\lambda\in{\mathbb{Z}}[A].$ Then $(M\otimes M)/\Phi(R)=M\otimes_{{\mathbb{Z}}[A]}M.$ ∎

The group $C=\langle t\rangle$ acts on ${\mathbb{Z}}/p[\![x]\!]$ by multiplication on $1-x.$ As above, we can extend the action of $C$ on ${\mathbb{Z}}[\![x]\!]$ to the action of $C\otimes{\mathbb{Z}}_{p}$ in a continuous way. The group

[TABLE]

is called the lamplighter group. We consider the ’double version’ of this group, double lamplighter group:

[TABLE]

Its pro-$p$-completion is equal to the semidirect product

[TABLE]

with the described above action of $C\otimes{\mathbb{Z}}_{p}$ on ${\mathbb{Z}}/p[\![x]\!]$ (see Proposition 4.12 of [8]). We call the group ${\mathcal{D}\mathcal{L}}$ double lamplighter pro-$p$-group.

Theorem 4.3**.**

The kernel of the comparison homomorphism for the double lamplighter pro-$p$-group

[TABLE]

is uncountable.

Proof.

For the sake of simplicity we set $H_{2}(-)=H_{2}(-,{\mathbb{Z}}/p)$ and $H_{2}^{\sf cont}(-)=H_{2}^{\sf cont}(-,{\mathbb{Z}}/p).$ Consider the homological spectral sequence $E$ of the short exact sequence ${\mathbb{Z}}/p[\![x]\!]^{2}\rightarrowtail{\mathcal{D}\mathcal{L}}\twoheadrightarrow C\otimes{\mathbb{Z}}_{p}.$ Then the zero line of the second page is trivial $E^{2}_{k,0}=H_{k}(C\otimes{\mathbb{Z}}_{p})=(\Lambda^{k}{\mathbb{Z}}_{p})\otimes{\mathbb{Z}}/p=0$ for $k\geq 2.$ Using Lemma 4.1 we obtain $H_{k}(C\otimes{\mathbb{Z}}_{p},{\mathbb{Z}}/p[\![x]\!])=H_{k}(C,{\mathbb{Z}}/p[C])=0$ for $k\geq 1,$ and hence $E^{2}_{k,1}=0$ for $k\geq 1.$ It follows that

[TABLE]

For any ${\mathbb{Z}}/p$-vector space $V$ Künneth formula gives a natural isomorphism

[TABLE]

Then we have a split monomorphism

[TABLE]

It easy to see that the groups ${\mathcal{D}\mathcal{L}}_{(i)}=((x^{i})\oplus(x^{i}))\rtimes(C\otimes p^{i}{\mathbb{Z}}_{p})$ form a fundamental system of open normal subgroups. Consider the quotients ${\mathcal{D}\mathcal{L}}^{(i)}={\mathcal{D}\mathcal{L}}/{\mathcal{D}\mathcal{L}}_{(i)}.$ Then

[TABLE]

The short exact sequence ${\mathbb{Z}}/p[\![x]\!]^{2}\rightarrowtail{\mathcal{D}\mathcal{L}}\twoheadrightarrow C\otimes{\mathbb{Z}}_{p}$ maps onto the short exact sequence $({\mathbb{Z}}/p[x]/(x^{i}))^{2}\rightarrowtail{\mathcal{D}\mathcal{L}}^{(i)}\twoheadrightarrow C/C^{p^{i}}.$ Consider the morphism of corresponding spectral sequences $E\to{}^{(i)}\!E.$ Using (4.1) we obtain

[TABLE]

Similarly to (4.2) we have a split monomorphism

[TABLE]

Then we need to prove that the kernel of the map

[TABLE]

is uncountable.

Consider the antipod $\sigma:{\mathbb{Z}}/p[C]\to{\mathbb{Z}}/p[C]$ i.e. the ring homomorphism given by $\sigma(t^{n})=t^{-n}.$ The antipod induces a homomorphism $\sigma:{\mathbb{Z}}/p[x]/(x^{i})\to{\mathbb{Z}}/p[x]/(x^{i})$ such that $\sigma(1-x)=1+x+x^{2}+\dots.$ It induces the continuous homomorphism $\sigma:{\mathbb{Z}}/p[\![x]\!]\to{\mathbb{Z}}/p[\![x]\!]$ such that $\sigma(x)=-x-x^{2}-\dots.$ Moreover, we consider the antipode $\sigma$ on ${\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}].$ Note that the homomorphisms

[TABLE]

commute with the antipodes.

By Lemma 4.2 the correspondence $a\otimes b\leftrightarrow a\otimes\sigma(b)$ gives isomorphisms

[TABLE]

[TABLE]

Moreover, since ${\mathbb{Z}}/p[C\otimes{\mathbb{Z}}_{p}]\to{\mathbb{Z}}/p[x]/(x^{i})$ is an epimorphism, we obtain

[TABLE]

Therefore the homomorphism (4.3) is isomorphic to the multiplication homomorphism

[TABLE]

whose kernel is uncountable by Proposition 3.3. ∎

5. Proof of Main Theorem

Since the double lamplighter pro-$p$-group is $3$-generated, we have a continuous epimorphism $\hat{F}_{p}\twoheadrightarrow{\mathcal{D}\mathcal{L}},$ where $F$ is the $3$-generated free group. Then the statement of the theorem for three generated free group follows from Proposition 4.3 and Corollary 2.6. Using that the $3$-generated free group is a retract of $k$-generated free group for $k\geq 3$, we obtain the result for $k\geq 3.$ The result for two generated free group follows from Lemma 11.2 of [3].

Acknowledgement

The research is supported by the Russian Science Foundation grant N 16-11-10073.