13
$\begingroup$

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
  • 5
    $\begingroup$ Hint: LERF implies residually finite. $\endgroup$
    – YCor
    Oct 24, 2014 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, 2014 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, 2014 at 20:10
  • 4
    $\begingroup$ 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, 2014 at 20:18

3 Answers 3

7
$\begingroup$

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

$\endgroup$
1
  • 3
    $\begingroup$ It is not LERF because of the Mikhailova subgroup with undecidable membership problem. $\endgroup$ Nov 3, 2014 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.

$\endgroup$
1
  • $\begingroup$ I find the result interesting: do you have a reference in mind? $\endgroup$
    – Seirios
    Oct 25, 2014 at 7:14
3
$\begingroup$

Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF.

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 $BS(2,3) = \langle a, b \mid b^{-1}a^2 b = a^3 \rangle$. 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.

$\endgroup$
4
  • 1
    $\begingroup$ I do not think that the second statement, starting with "Furthermore" is correct. Indeed, it is well-known that many metabelian lamplighter groups are LERF (see, for example, Proposition 3.19 in [de Cornulier, Yves "Finitely presented wreath products and double coset decompositions." Geom. Dedicata 122 (2006), 89–108.]), but they are not polycyclic. $\endgroup$ Mar 14, 2023 at 12:07
  • $\begingroup$ @AshotMinasyan Ah, yes, thanks, you're completely right -- the statement I make is true only for ERF (proved by Alperin) not LERF. $\endgroup$ Mar 14, 2023 at 15:41
  • $\begingroup$ "free-by-cyclic": you probably mean "(f.g. free)-by-cyclic", to infer residual finiteness? $\endgroup$
    – YCor
    Mar 14, 2023 at 15:57
  • 2
    $\begingroup$ @YCor All finitely generated (free-by-cyclic) groups are residually finite, not just (f.g. free)-by-cyclic. This is due to G. Baumslag, "Finitely generated cyclic extensions of free groups are residually finite", Bull. Amer. Math. Soc., 5, 87-94, 1971. $\endgroup$
    – ADL
    Mar 14, 2023 at 16:24

Your Answer

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