MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we have two finitely generated residually finite groups $G$ and $H$, is there are relation between

the profinite completions $\hat{G},\hat{H}$ and the profinite completion of a semidirect product $\hat{G \rtimes H}$

(and analogous question for pro-p completions)

share|cite|improve this question
Semidirect product may not be even residually finite. – Mark Sapir Apr 5 '11 at 23:27
Sorry , I forgot to say that the groups are finitely generated. – Mustafa Gokhan Benli Apr 5 '11 at 23:37
up vote 13 down vote accepted

Take a finite non-abelian simple group $A$ and consider the wreath product $G=A\wr \mathbb Z$. Let $N$ be any subgroup of finite index of $G$. Then $N\cap A^{\mathbb Z}\ne 1$. Let $g$ be a non-trivial element in the intersection. Suppose that the $i$-th coordinate $g_i$ of $g$ is not $1$. Since $A$ has trivial center, there exists $h\in A$ such that $[g_i,h]\ne 1$. Let $h'$ be the element of $A^{\mathbb Z}$ with $h$ on the $i$-th coordinate and trivial other coordinates. Then $[g,h']$ is in $N$ and has exactly one non-trivial coordinate (number $i$). Using the fact that $A$ is simple and the action of ${\mathbb Z}$ on $A^{\mathbb Z}$, we get that $N$ contains $A^{\mathbb Z}$. Hence the profinite (pro-p) completion of $G$ is the same as the profinite completion of $\mathbb Z$. Of course $G$ is a semidirect product of $A^{\mathbb Z}$ and $\mathbb Z$, both residually finite.

If $G, H$ are finitely generated, then $P=\hat G\rtimes \hat H=\hat{G\rtimes H}$. Indeed it is easy to see that the profinite completion of $G$ in $P$ is $\hat G$. That is because for every finite index subgroup $N$ of $G$ there exists a finite index subgroup $K$ in $G\rtimes H$ such that $K\cap G < N$.

share|cite|improve this answer
I think you want $N$ to be normal which you can assume of course. I am not sure about the second part. I think it is more complicated. I need to think about it a bit. – Yiftach Barnea Apr 6 '11 at 7:40
Isn't $G$ meant instead of $A$ in the sentence "Of course $A$ is a semidirect product of $A^{\mathbb {Z}}$ and $\mathbb{Z}$ ? – Ralph Apr 6 '11 at 8:01
@Ralph: yes, thanks. – Mark Sapir Apr 6 '11 at 8:18

I convinced myself the Mark's second claim is true. Here is a detailed argument. Let us start by checking whether $\widehat{G} \rtimes \widehat{H}$ actually exists. We assume that both $G$ and $H$ are is finitely generated. Let $\varphi:H \to \textrm{Aut}(G)$ be the map that defines the semidirect product.

First we need to check that $\varphi$ can be extended to $\textrm{Aut}(\widehat{G})$. As $G$ is finitely generated it has finetely many subgroups of index $n$, let $G_n$ be their interestion. Then $G_n$ is a characteristic subgroup of finite index in $G$. Moreover, every subgroup of finite index in $G$ contains one of the $G_n$'s. Thus, $\widehat{G}$ is the inverse limit of $G/G_n$. Now every autmorphism of $G$ preserves $G_n$, hence, $\textrm{Aut}(G)$ is embedded in $\textrm{Aut}(\widehat{G})$. We conclude that $\varphi$ can be extended to $\textrm{Aut}(\widehat{G})$.

We now need to recall the topology on $\textrm{Aut}(\widehat{G})$. The open neighborhoods of the identity are defined as $A(G_n)$ the kernel of the map from $\textrm{Aut}(\widehat{G})$ to $\textrm{Aut}(G/G_n)$. To extend $\varphi$ to $\widehat{H}$ we need $\varphi$ to be continuous on the profinite topology of $H$. Thus, we need that $H_n$ the kernel of the map from $H$ to $\textrm{Aut}(G/G_n)$ to be of finite index. If $H$ is finitely generated, tThis is indeed the case as $\textrm{Aut}(G/G_n)$ is a finite group. So $\varphi$ can be extended.

That means we can define $\widehat{G} \rtimes \widehat{H}$. Moreover, from the above argument $\varphi$ is continuous on $\widehat{H}$, so $\widehat{G} \rtimes \widehat{H}$ is a profinite group. We notice that $\widehat{G} \rtimes \widehat{H}$ is the inverse limit of $(G \rtimes H)/(G_n \rtimes N)$, where $n \in \mathbb{N}$ and $N$ is a normal subgroups of finite index in $H$.

We always have a map from the profinite completeion of a group onto any profinite completion with respect to some subgroups of finite index. So we get $\psi$ from $\widehat{G \rtimes H}$ onto $\widehat{G} \rtimes \widehat{H}$. Now, suppose $K$ is a subgroup of finite index in $G \rtimes H$. Let us look at $K \cap G$, it is a subgroup of finite index in $G$. Therefore, it contains some $G_n$. Also, $K \cap H$ is of finite index in $H$. Now, $G_n \rtimes (K \cap H)$ is a subgroup, it is of finite index in $G \rtimes H$, and it is contained in $K$. We deduce that that $\psi$ is an isomorphism.

Edit: I do not think it is necessary for $H$ to be finitely generated so I fixed the argument.

share|cite|improve this answer
Thanky you, this is a very nice answer. – Mustafa Gokhan Benli Apr 7 '11 at 4:47

Let $\mathscr{P}$ be any property such that whenever a group has $\mathscr{P}$ then all its subgroups also have $\mathscr{P}$. In [1] Theorem 3.1, K. W. Gruenberg has proved that if the wreath product $W= A \wr B$, is residually $\mathscr{P}$, then either $B$ is $\mathscr{P}$ or $A$ is abelian.

Consider $W= S_3 \wr \mathbb{Z}$, where $S_3$ is the symmetric group of degree 3. Since $S_3$ is not abelian, $\mathbb{Z}$ is not finite, and the subgroup of any finite group is finite, the group $W$ is not RF.

Clearly, $W= \prod_{i \in \mathbb{Z}} S_3 \rtimes \mathbb{Z}$, where $\mathbb{Z}$ and $\prod_{i \in \mathbb{Z}} S_3$ are residually finite.

[1] K. W. Gruenberg, Residual properties of infinite soluble groups}, Prec. London Math. Soc., Ser. 3, 7 (1957), 29--62.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.