Finitely presented not implies Noetherian
This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finitely presented group) need not satisfy the second group property (i.e., Noetherian group)
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- Finitely generated abelian is subgroup-closed, and hence, any finitely generated abelian group is Noetherian.
- Finitely generated nilpotent is subgroup-closed, and hence, any finitely generated nilpotent group is Noetherian.
- Equivalence of definitions of polycyclic group: A group is polycyclic if and only if it is Noetherian and solvable. Polycyclic groups are also finitely presented.
Example of the free group of rank two
Further information: Free group:F2
The free group of rank is finitely presented (it has a presentation with two generators and no relations) but it is not Noetherian. For instance, in the free group with freely generating set , the normal closure of is a free group on infinitely many generators: . Let's see why this is true.
Consider a free group on countably many generators and now consider an element that acts on these generators by the rule . Consider the semidirect product of the free group on the s with a cyclic group generated by . Clearly, this group is generated by and hence is a quotient of with . We see thus that and thus the normal closure of maps surjectively to the free group on the s. Since the mapping is bijective on a generating set, we see that in fact the normal closure of is free on a countable generating set.
Example of the Baumslag-Solitar group
Further information: Baumslag-Solitar group:BS(1,2)
This group, given by the presentation:
is finitely presented, but the normal closure of in it is not finitely generated, since it is isomorphic to the additive group of 2-adic rationals.
Note that gives a solvable group example of the non-implication. See also finitely presented and solvable not implies polycyclic.