3
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ No, $F_2\times F_2$ is not LERF (it is even not LPF: there are profinitely dense f.g. proper subgroups). $\endgroup$
    – YCor
    Nov 17, 2015 at 17:55
  • 2
    $\begingroup$ (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 $P\neq 1$. $\endgroup$
    – YCor
    Nov 17, 2015 at 18:13
  • 1
    $\begingroup$ 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. $\endgroup$
    – HJRW
    Nov 18, 2015 at 10:29
  • $\begingroup$ @HJRW Thanks! The answer to my question is negative as explained by Yves in his three comments. $\endgroup$
    – Pablo
    Nov 18, 2015 at 12:17

1 Answer 1

Reset to default
1
$\begingroup$

(converted from the comments) No, $F_2\times F_2$ is a counterexample, where $F_2$ is free on 2 generators.

Recall that a group is LPF if the profinite closure of every f.g. subgroup of infinite index has infinite index. This fails if there is a profinitely dense f.g. subgroup.

[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).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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