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A Noetherian group (also sometimes called slender groups) is a group for which every subgroup is finitely generated. (Equivalently, it satisfies the ascending chain condition on subgroups).

A finitely presented group is a group with a presentation that has finitely many generators and finitely many relations.

Flipping through some search results and references, I get the impression that there should be examples of Noetherian groups that are not finitely presented (because I can locate references to "finitely presented Noetherian group", a name that shouldn't exist if being Noetherian implies being finitely presented). However, I'm not able to get an explicit reference or example. I would be grateful if somebody could point out a reference or example.

For a solvable group, being Noetherian is equivalent to being polycyclic (i.e., having a subnormal series where all the successive quotients are cyclic groups), and polycyclic groups are finitely presented. Hence, any counterexample must be a non-solvable group.

[Note: My standard example of a finitely generated group that is not finitely presented is a wreath product of the group of integers with itself. But this is far from Noetherian.]

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  • $\begingroup$ Maybe it's worth pointing out that conversely, finitely presentable groups need not be Noetherian. For instance, the free group on 2 generators. So the two notions are incomparable. $\endgroup$
    – Tim Campion
    Commented Dec 13, 2014 at 17:17

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Tarski monsters provide examples of 2-generator noetherian groups that is not finitely presented.

Edit (YCor): Tarski monsters, as defined in the link (infinite groups of prime exponent $p$ in which every nontrivial proper is cyclic) exist for large $p$ and all currently known constructions of Tarski monsters are known to yield groups that are not finitely presented. However, it is unknown whether there exists a finitely presented Tarski monster.

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  • $\begingroup$ Maybe I'm being dense, but is it obvious the Tarski monster is not finitely presented? $\endgroup$
    – Steve D
    Commented May 26, 2010 at 22:35
  • $\begingroup$ Not really, I believe. This is a consequence of Theorem 26.3 of the book A. Yu. Ol'shanskii, Geometry of defining relations in groups, Kluwer Academic Publishers, 1991. Basically it follows from the construction of Tarski monsters. $\endgroup$
    – Primoz
    Commented May 26, 2010 at 22:46
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    $\begingroup$ Wikipedia says: "The construction of Ol'Shanskii shows in fact that there are continuum-many non-isomorphic Tarski Monster groups for each [large] prime". $\endgroup$
    – HJRW
    Commented May 26, 2010 at 22:47
  • $\begingroup$ @Primoz: Theorem 26.3 says that the relations $R=1, R\in\mathfrak{R}$ that were used in the construction are independent (that is $R=1$ does not follow from $R'=1, R'\in\mathfrak{R}\setminus\{R\}$). That tells us that we cannot use a finite subset of $\mathfrak{R}$. It does not say anything about other presentations. $\endgroup$
    – Johannes Hahn
    Commented May 26, 2010 at 22:53
  • $\begingroup$ ... and it does not tell anything about Tarski Monsters that were not constructed the way Ol'shanskii constructed them. Henry Wiltons point does provide a better way: there are only countably many finitely presented 2-generator groups because there are only countably many normal subgroups of the free group $F_2$ of rank 2. $\endgroup$
    – Johannes Hahn
    Commented May 26, 2010 at 22:56
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It's unknown whether every slender group is virtually polycyclic. See page 87 of Matt Clay's thesis.

EDIT: Primoz rightly points out that a Tarski monster is slender (and not finitely presentable!). This seems right. I'm not sure what to make of Clay's claim (which I'm fairly sure I've seen elsewhere). Presumably it's unknown whether there are finitely presented, non-virtually-polycyclic, slender groups. As James points out, one can impose other conditions, like residual finiteness, that rule out such pathological examples.

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  • $\begingroup$ Okay, so (unless I'm missing something obvious) if the statement is true (that every slender group is virtually polycyclic) then every slender group is also finitely presented. Are the statements strictly equivalent? i.e., is saying "every slender group is finitely presented" equivalent to saying "every slender group is virtually polycyclic" ? $\endgroup$
    – Vipul Naik
    Commented May 26, 2010 at 20:49
  • $\begingroup$ The last I heard (2009) this was unknown even for residually finite groups. (The question in this form is due to M.I. Kargapolov in the Kourovka notebook.) $\endgroup$
    – James
    Commented May 26, 2010 at 20:54
  • $\begingroup$ I was about to come and ask you about the Tarski monster, but Primoz already posted that as an answer. I guess the geometric constraints in Clay's thesis impose some condition such as "residual finiteness" which makes the problem open. $\endgroup$
    – Vipul Naik
    Commented May 27, 2010 at 17:24
  • $\begingroup$ Vipul, Yeah, I feel a little silly - of course I knew about Tarski monsters, but for some reason they were in different parts of my brain! If I see Matt around, I suppose I'll ask him. (I don't think residual finiteness is the right condition for him, though. Some of the groups he's interested in, such as Baumslag--Solitar groups, are certainly not residually finite.) $\endgroup$
    – HJRW
    Commented May 27, 2010 at 17:56

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