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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 on 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.

4 added 1 characters in body; edited body; added 4 characters in body

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 the 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 $G (G \rtimes H/G_n H)/(G_n \rtimes N$N)$, where$n \in \mathbb{N}$and$N$is a normal subgroups of finite index in$H$. We always have a map from the profinite completeion of a group on any profinite completion with respect to some subgroups of finite index. So we get$\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$to be finitely generated so I fixed the argument. 3 added 137 characters in body I convinced myself the Mark's second claim is true. Here is a detailed argument. Let us start by checking whether$\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)$be the map the defines the semidirect product. First we need to check that$\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$is finitely generated, this tThis is indeed the case as$\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$. We always have a map from the profinite completeion of a group on any profinite completion with respect to some subgroups of finite index. So we get$\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.

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