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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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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. –  Tim Campion Dec 13 at 17:17

2 Answers 2

up vote 8 down vote accepted

A Tarski monster is an example of a 2-generator noetherian group that is not finitely presented.

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Maybe I'm being dense, but is it obvious the Tarski monster is not finitely presented? –  Steve D May 26 '10 at 22:35
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. –  Primoz May 26 '10 at 22:46
Wikipedia says: "The construction of Ol'Shanskii shows in fact that there are continuum-many non-isomorphic Tarski Monster groups for each [large] prime". –  HJRW May 26 '10 at 22:47
Thank you both. –  Steve D May 26 '10 at 22:50
@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. –  Johannes Hahn May 26 '10 at 22:53

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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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" ? –  Vipul Naik May 26 '10 at 20:49
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.) –  James May 26 '10 at 20:54
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. –  Vipul Naik May 27 '10 at 17:24
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.) –  HJRW May 27 '10 at 17:56

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