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

Abstract

For a prime $p$, let $\hat{F}_{p}$ be a finitely generated free pro-$p$-group of rank at least $2$. 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.

1 Introduction

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

$$\begin{eqnarray}H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p),\end{eqnarray}$$

which connects discrete and continuous homology groups of  $G$ . Here $H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)=H_{2}(G,\mathbb{Z}/p)$ is the second homology group of $G$ with $\mathbb{Z}/p$ -coefficients, where $G$ is viewed as a discrete group. The continuous homology $H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)$ 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 $H_{2}^{\mathsf{disc}}\rightarrow H_{2}^{\mathsf{cont}}$ is the inverse limit of the coinflation maps $H_{2}(G,\mathbb{Z}/p)\rightarrow H_{2}(G/U,\mathbb{Z}/p)$ (see [FKRS08, Theorem 2.1]).

The study of the comparison map for different types of pro- $p$ -groups is a fundamental problem in the theory of profinite groups (see [FKRS08] for discussion and references). It is well known that for a finitely generated free pro- $p$ -group $\hat{F}_{p}$ ,

$$\begin{eqnarray}H_{2}^{\mathsf{cont}}(\hat{F}_{p},\mathbb{Z}/p)=0.\end{eqnarray}$$

Bousfield posed the following question in [Bou77, Problem 4.11], (case $R=\mathbb{Z}/n$ ).

Problem (Bousfield).

Does $H_{2}^{\mathsf{disc}}(\hat{F}_{n},\mathbb{Z}/n)$ 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 [Bou77, Proposition 12.3]). That is, the above problem is completely reduced to the case of homology groups $H_{2}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)$ for primes $p$ and, since $H_{2}^{\mathsf{cont}}(\hat{F}_{p},\mathbb{Z}/p)=0$ , the problem becomes a question about the non-triviality of the kernel of the comparison map for  $\hat{F}_{p}$ .

In [Bou92], Bousfield proved that, for a finitely generated free pro- $p$ -group $\hat{F}_{p}$ on at least two generators, the group $H_{i}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)$ 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 Bousfield–Kan sense.

The group $H_{2}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)$ plays a central role in the theory of $H\mathbb{Z}/p$ -localizations developed in [Bou77]. It follows immediately from the definition of $H\mathbb{Z}/p$ -localization that, for a free group  $F$ , $H_{2}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)=0$ if and only if $\hat{F}_{p}$ coincides with the $H\mathbb{Z}/p$ -localization of  $F$ . (From the point of view of profinite groups the Bousfield problem is also discussed in [Nik11, § 7] by Nikolov and in [Klo16, § 4] by Klopsch.)

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

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

There are two cases in Bousfield’s problem, $R=\mathbb{Z}/n$ and $R=\mathbb{Q}$ . We give the answer for the case of $R=\mathbb{Z}/n$ . (Recently the authors gave the solution for the $R=\mathbb{Q}$ case [IM17], using completely different methods.)

The proof is organized as follows. In § 2 we consider properties of discrete and continuous homology of profinite groups. Using a result of Nikolov and Segal [NS07, Theorem 1.4], we show that for a finitely generated profinite group $G$ and a closed normal subgroup $H$ the cokernels of the maps $H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{disc}}(G/H,\mathbb{Z}/p)$ and $H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{cont}}(G/H,\mathbb{Z}/p)$ coincide (Theorem 2.5):

As a corollary we obtain (Corollary 2.6) that, for a finitely generated free pro- $p$ -group $\hat{F}_{p}$ , a continuous epimorphism $\unicode[STIX]{x1D70B}:\hat{F}_{p}{\twoheadrightarrow}G$ to a pro- $p$ -group induces the exact sequence

(1.1) $$\begin{eqnarray}H_{2}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)\overset{\unicode[STIX]{x1D70B}_{\ast }}{\longrightarrow }H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)\overset{\unicode[STIX]{x1D711}}{\longrightarrow }H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)\longrightarrow 0.\end{eqnarray}$$

That is, to prove 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 non-zero kernel. Observe that the statements in § 2 significantly use the theory of profinite groups and there is no direct way to generalize them for pronilpotent groups. In particular, we do not 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 [Bou92]. 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 $C\,\otimes \,\mathbb{Z}_{p}=\{t^{\unicode[STIX]{x1D6FC}},\unicode[STIX]{x1D6FC}\in \mathbb{Z}_{p}\}$ . Consider the continuous multiplicative homomorphism $\unicode[STIX]{x1D70F}:C\,\otimes \,\mathbb{Z}_{p}\rightarrow \mathbb{Z}/p[\![x]\!]$ sending $t$ to $1-x$ . The main result of § 3 is Proposition 3.3, which claims that the kernel of the multiplication map

(1.2) $$\begin{eqnarray}\mathbb{Z}/p[\![x]\!]\otimes _{\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[\![x]\!]\longrightarrow \mathbb{Z}/p[\![x]\!]\end{eqnarray}$$

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 [Kro85]. The profinite completion of the $p$ -lamplighter group is considered in [GK14], where it is shown that it is a semi-similar group generated by finite automaton. We consider the double lamplighter group,

$$\begin{eqnarray}(\mathbb{Z}/p)^{2}\wr \mathbb{Z}=\langle a,b,c\mid [b,b^{a^{i}}]=[c,c^{a^{i}}]=[b,c^{a^{i}}]=b^{p}=c^{p}=1,\,i\in \mathbb{Z}\rangle .\end{eqnarray}$$

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 notation)

where the left vertical arrow is a split monomorphism and the upper horizontal map is the multiplication map (see proof of Theorem 4.3). This implies that, for the group ${\mathcal{D}}{\mathcal{L}}$ , the comparison map $H_{2}^{\mathsf{disc}}({\mathcal{D}}{\mathcal{L}},\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{cont}}({\mathcal{D}}{\mathcal{L}},\mathbb{Z}/p)$ 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 [Bou92, Lemma 11.2] to get the same result for a 2-generated free group  $F$ .

In [Bou77], Bousfield formulated the following generalization of the above problem for the class of finitely presented groups (see [Bou77, Problem 4.10], 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 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 $H_{2}(G,\mathbb{Z}/p)\rightarrow H_{2}({\hat{G}}_{p},\mathbb{Z}/p)$ ? It is shown in [Bou77] that this is true for the class of polycyclic groups. The same is true for finitely presented metabelian groups [IM16]. The 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 $H_{2}(P,\mathbb{Z}/p)\rightarrow H_{2}(\hat{P}_{p},\mathbb{Z}/p)$ 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 [NS07, Theorem 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 $H/[H,G]\rightarrow H/[H,G]$ 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 the fact that the preimage of a closed set under continuous function is closed, we obtain that $[H,G]\cdot H^{p}$ is closed.◻

Observe that, in the proof of Corollary 2.2, [NS07, Theorem 1.4] is not used in full generality. We only need it in the case of pro- $p$ groups, and in this particular case the proof of this theorem is quite elementary.

Lemma 2.3 (mod- $p$ Hopf formula).

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

$$\begin{eqnarray}H_{2}(G,\mathbb{Z}/p)\longrightarrow H_{2}(G/H,\mathbb{Z}/p)\longrightarrow \frac{H\cap ([G,G]G^{p})}{[H,G]H^{p}}\longrightarrow 0.\end{eqnarray}$$

Proof. This follows from the five-term exact sequence

$$\begin{eqnarray}H_{2}(G,\mathbb{Z}/p)\longrightarrow H_{2}(G/H,\mathbb{Z}/p)\longrightarrow H_{1}(H,\mathbb{Z}/p)_{G}\longrightarrow H_{1}(G,\mathbb{Z}/p)\end{eqnarray}$$

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 formula).

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

$$\begin{eqnarray}H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)\longrightarrow H_{2}^{\mathsf{cont}}(G/H,\mathbb{Z}/p)\longrightarrow \frac{H\cap \overline{([G,G]G^{p})}}{\overline{([H,G]H^{p})}}\longrightarrow 0.\end{eqnarray}$$

Proof. For the sake of simplicity we set $H_{\ast }(-)=H_{\ast }^{\mathsf{discr}}(-,\mathbb{Z}/p)$ and $H_{\ast }^{\mathsf{cont}}(-):=H_{\ast }^{\mathsf{cont}}(-,\mathbb{Z}/p)$ . Consider the five-term exact sequence [RZ00, Corollary 7.2.6]

$$\begin{eqnarray}H_{2}^{\mathsf{cont}}(G)\longrightarrow H_{2}^{\mathsf{cont}}(G)\longrightarrow H_{0}^{\mathsf{cont}}(G,H_{1}^{\mathsf{cont}}(H))\longrightarrow H_{1}^{\mathsf{cont}}(G).\end{eqnarray}$$

Continuous homology and cohomology of profinite groups are Pontryagin dual to each other [RZ00, Proposition 6.3.6]. There are isomorphisms

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

where $\text{Hom}$ denotes the set of continuous homomorphisms (see [Ser02, I.2.3]). It follows that $H_{1}^{\mathsf{cont}}(G)=G/\overline{[G,G]G^{p}}$ . Similarly, $H_{1}^{\mathsf{cont}}(H)=H/\overline{[H,H]H^{p}}$ . [RZ00, Lemma 6.3.3] implies that $H_{0}^{\mathsf{cont}}(G,M)=M/\overline{\langle m-mg\mid m\in M,g\in G\rangle }$ for any profinite $(\mathbb{Z}/p[G])^{\wedge }$ -module  $M$ . Therefore $H_{0}^{\mathsf{cont}}(G,H_{1}^{\mathsf{cont}}(H))=H/\overline{[H,H]H^{p}}$ . The assertion follows.◻

We denote by $\unicode[STIX]{x1D711}$ the comparison map

$$\begin{eqnarray}\unicode[STIX]{x1D711}:H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p).\end{eqnarray}$$

Theorem 2.5. Let $G$ be a finitely generated profinite group and $H$ a closed normal subgroup of  $G$ . Denote

$$\begin{eqnarray}\displaystyle Q^{\mathsf{disc}} & := & \displaystyle \mathsf{Coker}(H_{2}(G,\mathbb{Z}/p)\rightarrow H_{2}(G/H,\mathbb{Z}/p)),\nonumber\\ \displaystyle Q^{\mathsf{cont}} & := & \displaystyle \mathsf{Coker}(H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)\rightarrow H_{2}^{\mathsf{cont}}(G/H,\mathbb{Z}/p)).\nonumber\end{eqnarray}$$

Then the comparison maps $\unicode[STIX]{x1D711}$ induce an isomorphism $Q^{\mathsf{disc}}\cong Q^{\mathsf{cont}}$ :

Proof. This follows from Lemmas 2.3, 2.4 and Corollary 2.2. ◻

Corollary 2.6. Let $G$ be a finitely generated pro- $p$ -group and $\unicode[STIX]{x1D70B}:\hat{F}_{p}{\twoheadrightarrow}G$ be a continuous epimorphism from the pro- $p$ -completion of a finitely generated free group  $F$ . Then the sequence

$$\begin{eqnarray}H_{2}^{\mathsf{disc}}(\hat{F}_{p},\mathbb{Z}/p)\overset{\unicode[STIX]{x1D70B}_{\ast }}{\longrightarrow }H_{2}^{\mathsf{disc}}(G,\mathbb{Z}/p)\overset{\unicode[STIX]{x1D711}}{\longrightarrow }H_{2}^{\mathsf{cont}}(G,\mathbb{Z}/p)\longrightarrow 0\end{eqnarray}$$

is exact.

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

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

In this section we follow to ideas of Bousfield written in [Bou92, Lemmas 10.6, 10.7]. 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 $C\,\otimes \,\mathbb{Z}_{p}=\{t^{\unicode[STIX]{x1D6FC}}\mid \unicode[STIX]{x1D6FC}\in \mathbb{Z}_{p}\}$ ; $\mathbb{Z}/p[\![x]\!]$ is the ring of power series; $\mathbb{Z}/p(\!(x)\!)$ is the field of formal Laurent series.

Lemma 3.1. Let $\mathsf{A}$ be a subset of $\mathbb{Z}/p[\![x]\!]$ . Denote by $\mathsf{A}^{i}$ the image of $\mathsf{A}$ in $\mathbb{Z}/p[x]/(x^{p^{i}})$ . Assume that

$$\begin{eqnarray}\lim _{i\rightarrow \infty }|\mathsf{A}^{i}|/p^{p^{i}}=0.\end{eqnarray}$$

Then the interior of $\mathbb{Z}/p[\![x]\!]\setminus \mathsf{A}$ 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 $|\mathsf{A}^{s+i}|/p^{p^{i+s}}\leqslant p^{-p^{s}}$ . Then $|\mathsf{A}^{s+i}|\leqslant p^{p^{i}}$ . Hence the number of elements in $\mathsf{A}^{i+s}$ is less than the number of open sets $f+f_{t}+(x^{p^{s+i}})$ . It follows that there exists $t$ such that $\mathsf{A}\cap (f+f_{t}+(x^{p^{s+i}}))=\varnothing$ . The assertion follows.◻

Denote by

$$\begin{eqnarray}\unicode[STIX]{x1D70F}:C\otimes \mathbb{Z}_{p}\rightarrow \mathbb{Z}/p[\![x]\!]\end{eqnarray}$$

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  $\unicode[STIX]{x1D70F}$ . Then the degree of the extension $[\mathbb{Z}/p(\!(x)\!):K]$ is uncountable.

Proof. Denote the image of the map $\unicode[STIX]{x1D70F}:C\,\otimes \,\mathbb{Z}_{p}\rightarrow \mathbb{Z}/p[\![x]\!]$ by  $\mathsf{A}$ . Let $\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n})$ and $\unicode[STIX]{x1D6FD}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})$ , where $\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n},\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n}\in \mathbb{Z}/p,$ and $k\geqslant 1$ . Denote by $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}$ the subset of $\mathbb{Z}/p[\![x]\!]$ consisting of elements that can be written in the form

(3.1) $$\begin{eqnarray}\frac{\unicode[STIX]{x1D6FC}_{1}a_{1}+\cdots +\unicode[STIX]{x1D6FC}_{n}a_{n}}{\unicode[STIX]{x1D6FD}_{1}b_{1}+\cdots +\unicode[STIX]{x1D6FD}_{n}b_{n}},\end{eqnarray}$$

where $a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{n}\in \mathsf{A}$ and $\unicode[STIX]{x1D6FD}_{1}b_{1}+\cdots +\unicode[STIX]{x1D6FD}_{n}b_{n}\notin (x^{p^{k}})$ . Then $K\cap \mathbb{Z}/p[\![x]\!]=\bigcup _{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}$ .

Fix some $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k$ . Take $i\geqslant k$ and consider the images of $\mathsf{A}$ and $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}$ in $\mathbb{Z}/p[x]/(x^{p^{i}})$ . Denote them by $\mathsf{A}^{i}$ and $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}^{i}$ . Obviously $\mathsf{A}^{i}$ is the image of the map $C/C^{p^{i}}\rightarrow \mathbb{Z}/p[x]/(x^{p^{i}})$ that sends $t$ to $1-x$ . Then $\mathsf{A}^{i}$ consists of $p^{i}$ elements. Fix some elements $\bar{a}_{1},\ldots ,\bar{a}_{n},\bar{b}_{1},\ldots ,\bar{b}_{n}\in \mathsf{A}^{i}$ that have preimages $a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{n}\in \mathsf{A}$ such that the ratio (3.1) is in $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}$ . For any such preimages $a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{n}\in \mathsf{A}$ the image $\bar{r}$ of the ratio (3.1) satisfies the equation

$$\begin{eqnarray}\bar{r}\cdot (\unicode[STIX]{x1D6FD}_{1}\bar{b}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{n}\bar{b}_{n})=\unicode[STIX]{x1D6FC}_{1}\bar{a}_{1}+\cdots +\unicode[STIX]{x1D6FC}_{n}\bar{a}_{n}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D6FD}_{1}\bar{b}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{n}\bar{b}_{n}\notin (x^{p^{k}})$ , the annihilator of $\unicode[STIX]{x1D6FD}_{1}\bar{b}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{n}\bar{b}_{n}$ consists 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 $\bar{a}_{1},\ldots ,\bar{a}_{n},\bar{b}_{1},\ldots ,\bar{b}_{n}\in \mathsf{A}^{i}$ , and for any such variant there are no more than $p^{p^{k}}$ variants for the image of the ratio. Therefore

(3.2) $$\begin{eqnarray}|\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}^{i}|\leqslant p^{2in+p^{k}}.\end{eqnarray}$$

Take any sequence of elements $v_{1},v_{2},\ldots \in \mathbb{Z}/p(\!(x)\!)$ 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 $v_{1},v_{2},v_{3},\ldots$ by powers of  $x$ , we can assume that $v_{1},v_{2},\ldots \in \mathbb{Z}/p[\![x]\!]$ . Fix some $\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k$ as above. Set

$$\begin{eqnarray}\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}=\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}\cdot v_{1}+\cdots +\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k}\cdot v_{l}.\end{eqnarray}$$

Then $\sum _{m=1}^{\infty }Kv_{m}=\bigcup _{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l,j}\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}\cdot x^{-j}$ . Denote by $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}^{i}$ the image of $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}$ in $\mathbb{Z}/p[x]/(x^{i})$ . Then (3.2) implies $|\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}^{i}|\leqslant p^{(2in+p^{k})l}$ . Therefore

$$\begin{eqnarray}\lim _{i\rightarrow \infty }|\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}^{i}|/p^{p^{i}}=0.\end{eqnarray}$$

By Lemma 3.1 the interior of the complement of $\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}$ is dense in $\mathbb{Z}/p[\![x]\!]$ . By the Baire theorem $\sum _{m=1}^{\infty }Kv_{m}=\bigcup _{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l,j}\mathsf{A}_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},k,l}\cdot x^{-j}$ has empty interior. In particular, $\sum _{m=1}^{\infty }Kv_{m}\neq \mathbb{Z}/p(\!(x)\!)$ .◻

Proposition 3.3. Consider the ring homomorphism $\mathbb{Z}/p[C\,\otimes \,\mathbb{Z}_{p}]\rightarrow \mathbb{Z}/p[\![x]\!]$ induced by  $\unicode[STIX]{x1D70F}$ . Then the kernel of the multiplication map

(3.3) $$\begin{eqnarray}\mathbb{Z}/p[\![x]\!]\otimes _{\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[\![x]\!]\longrightarrow \mathbb{Z}/p[\![x]\!]\end{eqnarray}$$

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

$$\begin{eqnarray}\unicode[STIX]{x1D707}:\mathbb{Z}/p[\![x]\!]\otimes _{R}K\rightarrow \mathbb{Z}/p(\!(x)\!).\end{eqnarray}$$

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

$$\begin{eqnarray}\unicode[STIX]{x1D705}:\mathbb{Z}/p(\!(x)\!)\rightarrow \mathbb{Z}/p[\![x]\!]\otimes _{R}K\end{eqnarray}$$

given by

$$\begin{eqnarray}\unicode[STIX]{x1D705}\biggl(\mathop{\sum }_{i=-n}^{\infty }\unicode[STIX]{x1D6FC}_{i}x^{i}\biggr)=\mathop{\sum }_{i=0}^{\infty }\unicode[STIX]{x1D6FC}_{i+n}x^{i}\otimes x^{-n}.\end{eqnarray}$$

Since we have $ax\,\otimes \,b=a\,\otimes \,xb$ , $\unicode[STIX]{x1D705}$ does not depend on the choice of  $n$ , we just have to chose it big enough. Using this, we get that $\unicode[STIX]{x1D705}$ is well defined. Obviously $\unicode[STIX]{x1D707}\unicode[STIX]{x1D705}=\mathsf{id}$ . 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 non-trivial 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 $\unicode[STIX]{x1D705}\unicode[STIX]{x1D707}=\mathsf{id}$ . Therefore

(3.4) $$\begin{eqnarray}\mathbb{Z}/p[\![x]\!]\otimes _{R}K\cong \mathbb{Z}/p(\!(x)\!).\end{eqnarray}$$

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

(3.5) $$\begin{eqnarray}\mathbb{Z}/p[\![x]\!]\otimes _{R}\mathbb{Z}/p[\![x]\!]\longrightarrow \mathbb{Z}/p[\![x]\!]\end{eqnarray}$$

is uncountable.

For any ring homomorphism $R\rightarrow S$ 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

(3.6) $$\begin{eqnarray}\mathbb{Z}/p(\!(x)\!)\otimes _{K}\mathbb{Z}/p(\!(x)\!)\longrightarrow \mathbb{Z}/p(\!(x)\!).\end{eqnarray}$$

Assume to 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 countable-dimensional kernel. Finally, note that the homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{K}^{2}\mathbb{Z}/p(\!(x)\!)\rightarrow \mathbb{Z}/p(\!(x)\!)\otimes _{K}\mathbb{Z}/p(\!(x)\!)\end{eqnarray}$$

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 $\unicode[STIX]{x1D6EC}_{K}^{2}\mathbb{Z}/p(\!(x)\!)$ 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 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 ‘multiplicative’ notation $a^{\unicode[STIX]{x1D6FC}}:=a\otimes \unicode[STIX]{x1D6FC}$ for $a\in A$ and $\unicode[STIX]{x1D6FC}\in \mathbb{Z}_{p}$ . 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 [IM16], but we include 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

$$\begin{eqnarray}H_{\ast }(A,M)\cong H_{\ast }(A,\hat{M})\cong H_{\ast }(A\otimes \mathbb{Z}_{p},\hat{M}).\end{eqnarray}$$

Proof. The first isomorphism is proven in [BD75]. 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 $H_{\ast }(A,\hat{M})$ such that $A$ acts trivially on $H_{\ast }(A,\hat{M})$ . 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 [RZ00, Theorem 4.2.2]) and $A$ is dense in $A\,\otimes \,\mathbb{Z}_{p}$ , we obtain that the action of $A\,\otimes \,\mathbb{Z}_{p}$ on $H_{\ast }(C,\hat{M})$ is trivial. Note that $\mathbb{Z}_{p}/\mathbb{Z}$ is a divisible torsion free abelian group, and hence $A\,\otimes \,(\mathbb{Z}_{p}/\mathbb{Z})\cong \mathbb{Q}^{\oplus \mathbf{c}}$ , where $\mathbf{c}$ is the continuous cardinal. Then the second page of the spectral sequence of the short exact sequence $A\rightarrowtail A\,\otimes \,\mathbb{Z}_{p}{\twoheadrightarrow}\mathbb{Q}^{\oplus \mathbf{c}}$ with coefficients in $\hat{M}$ is $H_{n}(\mathbb{Q}^{\oplus \mathbf{c}},H_{m}(A,\hat{M}))$ , where $L_{m}:=H_{m}(A,\hat{M})$ is a trivial $\mathbb{Z}/p[\mathbb{Q}^{\oplus \mathbf{c}}]$ -module. Then by universal coefficient theorem we have

$$\begin{eqnarray}0\longrightarrow \unicode[STIX]{x1D6EC}^{n}(\mathbb{Q}^{\oplus \mathbf{c}})\otimes L_{m}\longrightarrow H_{n}(\mathbb{Q}^{\oplus \mathbf{c}},L_{m})\longrightarrow \mathsf{Tor}(\unicode[STIX]{x1D6EC}^{n-1}(\mathbb{Q}^{\oplus \mathbf{c}}),L_{m})\longrightarrow 0.\end{eqnarray}$$

Since $\unicode[STIX]{x1D6EC}^{n}(\mathbb{Q}^{\oplus \mathbf{c}})$ is torsion free and $L_{m}$ is a $\mathbb{Z}/p$ -vector space, we get $\unicode[STIX]{x1D6EC}^{n}(\mathbb{Q}^{\oplus \mathbf{c}})\otimes L_{m}=0$ and $\mathsf{Tor}(\unicode[STIX]{x1D6EC}^{n-1}(\mathbb{Q}^{\oplus \mathbf{c}}),L_{m})=0.$ It follows that $H_{n}(\mathbb{Q}^{\oplus \mathbf{c}},L_{m})=0$ for $n\geqslant 1$ and $H_{0}(\mathbb{Q}^{\oplus \mathbf{c}},L_{m})=L_{m}$ . Then the spectral sequence consists of only one column, and hence $H_{\ast }(A\otimes \mathbb{Z}_{p},\hat{M})=H_{\ast }(A,\hat{M})$ .◻

Lemma 4.2. Let $A$ be an abelian group, $M$ be a $\mathbb{Z}[A]$ -module and $\unicode[STIX]{x1D70E}_{M}:M\rightarrow M$ be an automorphism of the underlying abelian group such that $\unicode[STIX]{x1D70E}_{M}(ma)=\unicode[STIX]{x1D70E}_{M}(m)a^{-1}$ for any $m\in M$ and $a\in A$ . Then there is an isomorphism

$$\begin{eqnarray}(M\otimes M)_{A}\cong M\otimes _{\mathbb{Z}[A]}M\end{eqnarray}$$

given by

$$\begin{eqnarray}m\otimes m^{\prime }\leftrightarrow m\otimes \unicode[STIX]{x1D70E}_{M}(m^{\prime }).\end{eqnarray}$$

Proof. Consider the isomorphism $\unicode[STIX]{x1D6F7}:M\,\otimes \,M\rightarrow M\,\otimes \,M$ given by $\unicode[STIX]{x1D6F7}(m\,\otimes \,m^{\prime })=m\,\otimes \,\unicode[STIX]{x1D70E}(m^{\prime })$ . 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 $\unicode[STIX]{x1D6F7}(R)$ is generated by $ma\,\otimes \,\unicode[STIX]{x1D70E}_{M}(m^{\prime })-m\,\otimes \,\unicode[STIX]{x1D70E}_{M}(m^{\prime })a$ . Using the fact that $\unicode[STIX]{x1D70E}_{M}$ 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 $\unicode[STIX]{x1D6F7}(R)$ , we obtain that $\unicode[STIX]{x1D6F7}(R)$ is generated by elements $m\unicode[STIX]{x1D706}\,\otimes \,m^{\prime }-m\,\otimes \,m^{\prime }\unicode[STIX]{x1D706}$ , where $\unicode[STIX]{x1D706}\in \mathbb{Z}[A]$ . Then $(M\,\otimes \,M)/\unicode[STIX]{x1D6F7}(R)=M\,\otimes _{\mathbb{Z}[A]}\,M$ .◻

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

$$\begin{eqnarray}\mathbb{Z}/p\wr C=\mathbb{Z}/p[C]\rtimes C=\langle a,b\mid [b,b^{a^{i}}]=b^{p}=1,i\in \mathbb{Z}\rangle\end{eqnarray}$$

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

$$\begin{eqnarray}(\mathbb{Z}/p[C]\oplus \mathbb{Z}/p[C])\rtimes C=\langle a,b,c\mid [b,b^{a^{i}}]=[c,c^{a^{i}}]=[b,c^{a^{i}}]=b^{p}=c^{p}=1,i\in \mathbb{Z}\rangle .\end{eqnarray}$$

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

$$\begin{eqnarray}{\mathcal{D}}{\mathcal{L}}=(\mathbb{Z}/p[\![x]\!]\oplus \mathbb{Z}/p[\![x]\!])\rtimes (C\otimes \mathbb{Z}_{p}),\end{eqnarray}$$

with the action of $C\otimes \mathbb{Z}_{p}$ on $\mathbb{Z}/p[\![x]\!]$ described above (see [IM16, Proposition 4.12]). We call the group ${\mathcal{D}}{\mathcal{L}}$ the double lamplighter pro- $p$ -group.

Theorem 4.3. The kernel of the comparison homomorphism for the double lamplighter pro- $p$ -group,

$$\begin{eqnarray}\unicode[STIX]{x1D711}:H_{2}^{\mathsf{disc}}({\mathcal{D}}{\mathcal{L}},\mathbb{Z}/p)\longrightarrow H_{2}^{\mathsf{cont}}({\mathcal{D}}{\mathcal{L}},\mathbb{Z}/p),\end{eqnarray}$$

is uncountable.

Proof. For the sake of simplicity we set $H_{2}(-)=H_{2}(-,\mathbb{Z}/p)$ and $H_{2}^{\mathsf{cont}}(-)=H_{2}^{\mathsf{cont}}(-,\mathbb{Z}/p)$ . 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: $E_{k,0}^{2}=H_{k}(C\otimes \mathbb{Z}_{p})=(\unicode[STIX]{x1D6EC}^{k}\mathbb{Z}_{p})\otimes \mathbb{Z}/p=0$ for $k\geqslant 2$ . 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 $k\geqslant 1$ , and hence $E_{k,1}^{2}=0$ for $k\geqslant 1$ . It follows that

(4.1) $$\begin{eqnarray}H_{2}({\mathcal{D}}{\mathcal{L}})=E_{0,2}^{2}.\end{eqnarray}$$

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

$$\begin{eqnarray}H_{2}(V\oplus V)\cong (V\otimes V)\oplus H_{2}(V)^{2}.\end{eqnarray}$$

Then we have a split monomorphism,

(4.2) $$\begin{eqnarray}(\mathbb{Z}/p[\![x]\!]\otimes \mathbb{Z}/p[\![x]\!])_{C\otimes \mathbb{Z}_{p}}\rightarrowtail E_{0,2}^{2}=H_{2}({\mathcal{D}}{\mathcal{L}}).\end{eqnarray}$$

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

$$\begin{eqnarray}H_{2}^{\mathsf{cont}}({\mathcal{D}}{\mathcal{L}})=\varprojlim H_{2}({\mathcal{D}}{\mathcal{L}}^{(i)}).\end{eqnarray}$$

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 $E\rightarrow ^{(i)}\!E$ . Using (4.1), we obtain

$$\begin{eqnarray}\text{Ker}(H_{2}({\mathcal{D}}{\mathcal{L}})\rightarrow H_{2}({\mathcal{D}}{\mathcal{L}}^{(i)}))\supseteq \text{Ker}(E_{2,0}^{2}\rightarrow ^{(i)}\!E_{2,0}^{2}).\end{eqnarray}$$

Similarly to (4.2), we have a split monomorphism

$$\begin{eqnarray}(\mathbb{Z}/p[x]/(x^{i})\otimes \mathbb{Z}/p[x]/(x^{i}))_{C\otimes \mathbb{Z}_{p}}\rightarrowtail ^{(i)}E_{2,0}^{2}.\end{eqnarray}$$

Then we need to prove that the kernel of the map

(4.3) $$\begin{eqnarray}(\mathbb{Z}/p[\![x]\!]\otimes \mathbb{Z}/p[\![x]\!])_{C\otimes \mathbb{Z}_{p}}\longrightarrow \varprojlim (\mathbb{Z}/p[x]/(x^{i})\otimes \mathbb{Z}/p[x]/(x^{i}))_{C\otimes \mathbb{Z}_{p}}\end{eqnarray}$$

is uncountable.

Consider the antipod $\unicode[STIX]{x1D70E}:\mathbb{Z}/p[C]\rightarrow \mathbb{Z}/p[C]$ , that is, the ring homomorphism given by $\unicode[STIX]{x1D70E}(t^{n})=t^{-n}$ . The antipod induces a homomorphism $\unicode[STIX]{x1D70E}:\mathbb{Z}/p[x]/(x^{i})\rightarrow \mathbb{Z}/p[x]/(x^{i})$ such that $\unicode[STIX]{x1D70E}(1-x)=1+x+x^{2}+\cdots \,$ . It induces the continuous homomorphism $\unicode[STIX]{x1D70E}:\mathbb{Z}/p[\![x]\!]\rightarrow \mathbb{Z}/p[\![x]\!]$ such that $\unicode[STIX]{x1D70E}(x)=-x-x^{2}-\cdots \,$ . Moreover, we consider the antipode $\unicode[STIX]{x1D70E}$ on $\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]$ . Note that the homomorphisms

$$\begin{eqnarray}\mathbb{Z}/p[C]\rightarrow \mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]\rightarrow \mathbb{Z}/p[\![x]\!]\rightarrow \mathbb{Z}/p[x]/(x^{i})\end{eqnarray}$$

commute with the antipodes.

By Lemma 4.2 the correspondence $a\otimes b\leftrightarrow a\otimes \unicode[STIX]{x1D70E}(b)$ gives isomorphisms

$$\begin{eqnarray}\displaystyle (\mathbb{Z}/p[\![x]\!]\otimes \mathbb{Z}/p[\![x]\!])_{C\otimes \mathbb{Z}_{p}} & \cong & \displaystyle \mathbb{Z}/p[\![x]\!]\otimes _{\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[\![x]\!],\nonumber\\ \displaystyle (\mathbb{Z}/p[x]/(x^{i})\otimes \mathbb{Z}/p[x]/(x^{i}))_{C\otimes \mathbb{Z}_{p}} & \cong & \displaystyle \mathbb{Z}/p[x]/(x^{i})\otimes _{\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[x]/(x^{i}).\nonumber\end{eqnarray}$$

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

$$\begin{eqnarray}\mathbb{Z}/p[x]/(x^{i})\otimes _{\mathbb{ Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[x]/(x^{i})\cong \mathbb{Z}/p[x]/(x^{i}).\end{eqnarray}$$

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

$$\begin{eqnarray}\mathbb{Z}/p[\![x]\!]\otimes _{\mathbb{Z}/p[C\otimes \mathbb{Z}_{p}]}\mathbb{Z}/p[\![x]\!]\longrightarrow \mathbb{Z}/p[\![x]\!],\end{eqnarray}$$

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 the 3-generated free group follows from Proposition 4.3 and Corollary 2.6. Using the fact that the 3-generated free group is a retract of the $k$ -generated free group for $k\geqslant 3$ , we obtain the result for $k\geqslant 3$ . The result for the 2-generated free group follows from [Bou92, Lemma 11.2].