# Is a finitely generated residually free group "almost LERF"?

Let $G$ be a finitely generated residually free group.

(i.e. for each $1 \neq g \in G$ there exists a homomorphism $\tau \colon G \to F$ such that $F$ is a free group, and $\tau(g) \neq 1$.)

Let $H \lneq G$ be a proper finitely generated subgroup. Must there be a finite index proper subgroup $U$ of $G$ containing $H$?

If $G$ must be LERF then the answer is positive. Henry Wilton, in a beautiful work, has proved that limit groups (fully residually free groups) are LERF.

• No, $F_2\times F_2$ is not LERF (it is even not LPF: there are profinitely dense f.g. proper subgroups).
– YCor
Nov 17, 2015 at 17:55
• (More precisely non-LPF means that there are f.g. subgroups of infinite index with profinite closure of finite index; LPF is weaker than LERF.) Now if $F$ is f.g. free, $P$ is a f.p. quotient group of $F$ with no proper finite quotient, then the fiber product $F\times_P F$ is f.g., and profinitely dense in $F\times F$, and of infinite index as soon as $P\neq 1$.
– YCor
Nov 17, 2015 at 18:13
• You might be interested in my paper with Martin Bridson: arxiv.org/abs/0706.4247 . We prove that every finitely presented subgroup of a residually free group is separable. As well as my result about limit groups, the extra ingredient is a theorem of Bridson--Howie--Miller--Short, and the answer to your question might be in their work.
– HJRW
Nov 18, 2015 at 10:29
• @HJRW Thanks! The answer to my question is negative as explained by Yves in his three comments. Nov 18, 2015 at 12:17

(converted from the comments) No, $F_2\times F_2$ is a counterexample, where $F_2$ is free on 2 generators.
[Remark: your property appeared in the 3-manifold literature as "$G$ has the engulfing property". It is an elementary remark (see page 10-11 here, where Property LPF is introduced) that a group has Property LPF iff each of its finite index subgroups has the engulfing property.]
Now a way to get profinitely dense f.g. subgroups in $F_2\times F_2$ is as follows. Consider an aperiodic (= with no nontrivial finite quotient) infinite finitely presented group $P$ on 2 generators (or with $n$ generators, but then work with $F_n\times F_n$), and fix an epimorphism $f:F_2\to P$. Then the fibre product $$F_2\times_P F_2=\{(x,y)\in F_2\times F_2:f(x)=f(y)\}$$ is profinitely dense in $F_2\times F_2$ (because $P$ is aperiodic) and finitely presented (because $P$ is finitely presented) and has infinite index (because $P$ is infinite).