Return to Answer

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If you do not $K$ to be trivial, then take $G={\mathbb Z}\wr C_2$, $K=C_2$, you get a trivial $\mathcal F$. On the other hand take $G={\mathbb Z} * C_2$ and represent it as $N\rtimes C_2$.

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If you do not want $K$ to be trivial, then take $G={\mathbb Z}\wr C_2$, $K=C_2$, you get a trivial $\mathcal F$. On the other hand take $G={\mathbb Z} * C_2$ and represent it as $N\rtimes C_2$.

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If you do not $K$ to be trivial, then take $G={\mathbb Z}\wr C_2$, $K=C_2$, you get a trivial $\mathcal F$. On the other hand take $G={\mathbb Z} * C_2$ and represent it as $N\rtimes C_2$.

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the second summand, $N$ is the first summand. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense.

Take $G={\mathbb Z}^2={\mathbb Z}\times {\mathbb{Z}}$, $K$ be the trivial group, $N=G$. Then $G$ is finitely generated, $\mathcal F$ is trivial and not profinitely dense. On the other hand if $G$ is the free group of rank $2$ and we represent $G=G\rtimes K$ where $K$ is the trivial group, $N=G$, then $\mathcal F=G$ is profinitely dense. If