# Example of Noetherian group (every subgroup is finitely generated) that is not finitely presented

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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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