Are all subgroups of a LERF group closed? A group $G$ is LERF if every finitely generated subgroup of G is closed in the profinite topology of $G$.
Let $G$ be LERF group, and let $H$ be a subgroup of $G$. Is $H$ necessarily closed? (I'm not assuming that $H$ is finitely  generated).
What if we assume that $G$ itself is finitely generated, or even finitely presented?
 A: Let $F$ be a finitely generated free group. It is known that $F$ is LERF. Now, let $N  \lhd F$ be a normal subgroup. If $N$ is closed (with respect to the profinite topology), then, for every $g \in F \backslash N$, there exists a finite-index subgroup $H \leq F$ containing $N$ but not containing $g$; ie., the subgroup $N$ is separable. This condition is equivalent to: the quotient $G/N$ is residually finite. 
On the other hand, it is known that the Baumslag-Solitar group $BS(2,3)= \langle a,b \mid ba^2b^{-1}=a^3 \rangle$ is not residually finite (in fact, it is not hopfian). 
Thus, the normal subgroup $\langle \langle ba^2b^{-1}a^{-3} \rangle \rangle$ cannot be closed in the free group $\langle a,b \mid \ \rangle$.
A: A group $G$ is called $ERF$ if every subgroup of $G$ is closed in the profinite ropology of $G$.
You're are asking if $LERF \Rightarrow ERF$, if I'm understanding you correctly. However this is not true, see e.g. https://mathoverflow.net/questions/220069/profinite-topology-on-free-metabelian-group.
Moreover see Robinson, Russo, Vincenzi "On groups whose subgroups are closed in the profinite topology", which studies the property $ERF$ more thoroughly.
