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Math-1820

Questions in Geometric Group Theory — Q 13.4

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Statement

Definitions and mathematical statement

Call a group $G$ slender in the source's group-theoretic sense if every subgroup of $G$ is finitely generated. This is also called the Noetherian condition on subgroups; it is unrelated to the different use of slenderness for homomorphisms from infinite direct products in abelian-group theory.

A group $P$ is polycyclic if it has a finite subnormal series \[ 1=P_0\trianglelefteq P_1\trianglelefteq\cdots \trianglelefteq P_r=P \] whose successive quotients $P_i/P_{i-1}$ are cyclic, possibly finite. A group is polycyclic-by-finite if it contains a polycyclic normal subgroup of finite index; equivalently, it is virtually polycyclic.

Does there exist a finitely presented slender group which is not polycyclic-by-finite? Finitely presented means that the group can be specified by finitely many generators and finitely many defining relations. Every subgroup must satisfy finite generation, with no uniform bound imposed on the sizes of generating sets.

The requirement applies to arbitrary subgroups, not merely normal subgroups or finite-index subgroups. The sought group itself must have a finite presentation; examples known only through infinitely many relations would not meet this condition. A group containing a nonabelian free subgroup cannot qualify, since a free group of rank two has infinitely generated subgroups. The problem nevertheless asks for the full existence statement rather than adding solvability, linearity, torsion-freeness, or a particular geometric action as further restrictions.

Short English statement

Is there a finitely presented group all of whose subgroups are finitely generated, yet which is not virtually polycyclic?

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