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Do there exist two elementary equivalent finitely generated groups $G,H$ such that $G$ is finitely presented but $H$ is not finitely presentable?

It seems reasonable to think that finite presentability is not preserved under elementary equivalence among finitely generated groups, so such groups probably exist. However, I do not know any example. The case where $G,H$ have the same universal theory, instead of having the same first order theory, would be already interesting.

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    $\begingroup$ @TimCampion of course not: an infinite f.g. object is EE to its ultrapowers, which have cardinal continuum. $\endgroup$
    – YCor
    Dec 30, 2021 at 17:24
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    $\begingroup$ @YCor Oh sure, wow. Alternatively, just consider upward Lowenheim-Skolem... But I wonder whether a countable group EE to a fg group must be fg... $\endgroup$
    – Tim Campion
    Dec 30, 2021 at 17:32
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    $\begingroup$ @TimCampion no, fg is not EE-invariant among countable groups: $\mathbf{Z}$ and $\mathbf{Z}\times\mathbf{Q}$ are EE. (I even guess that every infinite group is EE to a countable group that is not finitely generated.) $\endgroup$
    – YCor
    Dec 30, 2021 at 19:02
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    $\begingroup$ @BenjaminSteinberg not this previous example, but however here's such an easy formula (namely, that is true for some countable group but for no f.g. group): "($G$ is abelian) and (every element of $G$ is a square) and (there exists an element of order 2 in $G$)". $\endgroup$
    – YCor
    Dec 30, 2021 at 19:06
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    $\begingroup$ @TimCampion Any infinite group is elementarily equivalent to a countable group that is not f.g. First, make the group countable by downward LS. Repeatedly using the fact that any infinite countable structure has a countable proper elementary extension, build a strictly increasing elementary $\omega$-chain of countable elementary extensions of the group. The union of the chain is then a countable elementary extension, and it is not f.g., as any finite subset of the union is included in one of the groups in the chain. $\endgroup$ Dec 31, 2021 at 7:51

3 Answers 3

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The "universal theory" question is easy to solve: indeed if $G$ is a group and $H$ a subgroup, any universal formula true for $G$ is true for $H$. Hence if $G,H$ are groups and both embed into each other, then $G,H$ have the same universal theory.

Denote by $F_2$ the free group on two generators.

Now let $G$ be $F_2\times F_2$ and $H$ the kernel of the homomorphism $G\to\mathbf{Z}$ mapping all four generators to $1$. Then $H$ is not finitely presented, and contains a copy of $G$. So $G,H$ have the same universal theory.

(However they are not EE since $H$ doesn't satisfy the formula expressing: there are elements $x_1,\dots,x_4$ such that each element is uniquely the product of an element in the centralizer of $\{x_1,x_2\}$ and one in the centralizer of $\{x_3,x_4\}$.)

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  • $\begingroup$ Please clarify what $F_2$ is. $\endgroup$ Jan 1 at 12:39
  • $\begingroup$ @PaulTaylor, it's a free group on 2 generators $\endgroup$ Jan 1 at 20:28
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The following is not an answer to the main question, but provides some context which any answer may need to take into account. This context is certainly well known to many participants in the discussion, but making it explicit may be beneficial.

As the OP says, it is reasonable to conjecture that there is a finitely presented group $G$ and a finitely generated, but infinitely presented, group $H$, that are elementarily equivalent. But confirming this conjecture may be extremely difficult, for the simple reason that it is often extremely difficult to confirm that any pair of finitely generated groups is elementarily equivalent.

The key problem in this area was a famous question of Tarski:

Question (Tarski, c. 1945): Is the free group on 2 generators, $F_2$, elementarily equivalent to the free group on 3 generators, $F_3$?

Tarski's question was answered affirmatively by Sela around 2000, but can be seen to be very difficult by at least two different measures. First, it took more than 50 years to answer. Second, Sela's solution spans seven papers and many hundreds, or even thousands, of pages of mathematics. In the subseqent 20 years, the community has been unable to provide any significant simplification of Sela's proof.

So an answer to this question may be as difficult as Sela's proof, or even more so. Sela's work extends to all torsion-free hyperbolic groups $\Gamma$, and indeed he is able to classify all finitely generated groups $G$ elementarily equivalent to such $\Gamma$. Unfortunately for this question, he proved that any such $G$ is also hyperbolic, in particular finitely presented. The work of Dahmani, Groves, Guirardel, Hull, Reinfeldt and Weidmann in various combinations begins to extend Sela's techniques to more general classes of "negatively curved" groups, but I think full proofs of elementary equivalence for any of these groups are still a long way off.

In summary, this question may well be extremely difficult, and my guess is that it's wide open. If the question is important I would ask Sela himself, and regard his answer as definitive.

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  • $\begingroup$ Now that the theory of nonabelian free groups has also been proved decidable, is there a nice axiomatization for it? $\endgroup$
    – Matt F.
    Jan 6 at 10:23
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    $\begingroup$ @MattF.: In a word, no. :) Kharlampovich--Myasnikov's proof of decidability is also exceptionally long and complicated, and doesn't really offer any structural insights beyond Sela's theorems. (Actually, I should probably also mention that there is some controversy about whether or not the KM proof of decidability is correct. Some people still regard decidability as an open problem.) $\endgroup$
    – HJRW
    Jan 6 at 16:40
  • $\begingroup$ No axiomatization after 15 years is worrisome, since extracting one from a proof of decidability is usually straightforward. Also I now see Sela's report that he used a non-effective quantifier elimination (!) and "the decidability of the theories of both free and hyperbolic groups should be considered as open problems". (That's from the short version ma.huji.ac.il/~zlil/decidability.pdf, with a longer version at math.huji.ac.il/~zlil/km.pdf.) $\endgroup$
    – Matt F.
    Jan 6 at 18:42
  • $\begingroup$ @MattF.: Even when/if the controversy has been resolved and the proofs have been digested, I doubt that it will be reasonable to hope for a "nice axiomatisation". The bottom line is that the elementary theory of free groups is extraordinarily complicated. Apparently Razborov gave up on Tarski's problem after he found an example of a sentence that isn't reducible to an AE sentence. IIRC, Sela says he's glad Razborov never told him about this, or he would have given up too! (Sela's proof reduces quantifiers to boolean combinations of AE sentences.) $\endgroup$
    – HJRW
    Jan 7 at 17:32
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The problem is known and still open but does not seem hopeless. I would start with considering Zilber's old example of two f.g. non-isomorphic nilpotent of class $2$ groups which are e.e.: B.I. Zilʹber, An example of two elementarily equivalent but not isomorphic finitely generated metabelian groups. Algebra i Logika 10 (1971), 309–315. Both of his groups were of course finitely presented. But I would guess that there are similar examples when one of the groups is not finitely presented.

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