10
$\begingroup$

A finitely generated group $G$ is called LERF if every finitely generated $H \leq G$ is closed in the profinite topology on $G$ (equivalently, there is a family of finite index subgroups of $G$ intersecting in $H$). I am interested in examples of families of groups which are known to be LERF. In view of Marshall Hall's theorem for surface groups some questions come to mind:

Is every one-relator group LERF? (NO, since by YCor there are such groups which are not even residually finite).

Which finitely presented groups are LERF?

Which groups are known to be residually finite but not LERF?

$\endgroup$
  • 4
    $\begingroup$ Hint: LERF implies residually finite. $\endgroup$ – YCor Oct 24 '14 at 10:26
  • 1
    $\begingroup$ Some have already voted to close this as too broad. The problem is really the question 'Which finitely presented groups are LERF?' As with most questions about finitely presented groups, the answer is that LERF is not a recursive property, and so no classification is possible. A slightly more subtle question is whether LERF is a recursively enumerable property - I doubt it, but a proof that it isn't doesn't spring to mind right now. $\endgroup$ – HJRW Oct 24 '14 at 20:04
  • 1
    $\begingroup$ Actually, on a related note, I don't think I know a proof that the set of residually finite fp groups isn't recursively enumerable. $\endgroup$ – HJRW Oct 24 '14 at 20:10
  • 2
    $\begingroup$ One more thing: 'Marshall Hall's theorem' for surface groups is not Marshall Hall's theorem, it's Peter Scott's theorem. $\endgroup$ – HJRW Oct 24 '14 at 20:18
5
$\begingroup$

The group $F_2 \times F_2$ is not LERF, by Allenby-Gregorac 1973, which has lots of other results in this vein.

$\endgroup$
  • 3
    $\begingroup$ It is not LERF because of the Mikhailova subgroup with undecidable membership problem. $\endgroup$ – Anton Klyachko Nov 3 '14 at 2:00
5
$\begingroup$

A right-angled artin group is lerf iff its defining graph contains no induced subgraph which is a square or a path with 4 vertices. The path with 4 vertices is interesting because it is not lerf by Niblo-Wise but it has a decidable genetalized word problem, which is necessary for a finitely presented lerf group.

$\endgroup$
  • $\begingroup$ I find the result interesting: do you have a reference in mind? $\endgroup$ – Seirios Oct 25 '14 at 7:14
1
$\begingroup$

Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF. Furthermore, it is known that a solvable group is LERF if and only if it is polycyclic.

As mentioned in the comments, as not all one-relator groups are residually finite, not all one-relator groups are LERF. An example is, of course, the Baumslag-Solitar group $BS(2,3) = \langle a, b \mid b^{-1}ab = a^2 \rangle$. An example of a residually finite one-relator group which is not LERF is the Burns-Karrass-Solitar group $\langle a, b \mid [a, a^b] = 1 \rangle$. This contains $A(P_4)$, as in Benjamin Steinberg's answer, and is hence not LERF. It is however residually finite -- indeed, it is even free-by-cyclic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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