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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Statement

A finitely presented group (i.e., a group that admits a finite presentation) need not be a Noetherian group (also called slender group: a group in which every subgroup is finitely generated).

Related facts

Converse

• Noetherian not implies finitely presented

Similar facts

• Finitely presented and solvable not implies Noetherian

Opposite facts

• 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.

Proof

Example of the free group of rank two

Further information: Free group:F2

The free group of rank $2$ is finitely presented (it has a presentation with two generators and no relations) but it is not Noetherian. For instance, in the free group $\langle a,b\rangle$ with freely generating set $\{a,b\}$, the normal closure of $\langle a\rangle$ is a free group on infinitely many generators: $\{b^{n}ab^{-n}\mid n\in \mathbb {Z} \}$. Let's see why this is true.

Consider a free group on countably many generators $x_{n},n\in \mathbb {Z}$ and now consider an element $y$ that acts on these generators by the rule $x_{n}\mapsto x_{n+1}$. Consider the semidirect product of the free group on the $x_{n}$s with a cyclic group generated by $y$. Clearly, this group is generated by $x_{0},y$ and hence is a quotient of $\langle a,b\rangle$ with $a\mapsto x_{0},b\mapsto y$. We see thus that $b^{n}ab^{-n}\mapsto x_{n}$ and thus the normal closure of $\langle a\rangle$ maps surjectively to the free group on the $x_{n}$s. Since the mapping is bijective on a generating set, we see that in fact the normal closure of $\langle a\rangle$ 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:

$\langle a,b\mid bab^{-1}=a^{2}\rangle$

is finitely presented, but the normal closure of $\langle a\rangle$ in it is not finitely generated, since it is isomorphic to the additive group $\mathbb {Z} [1/2]$ of 2-adic rationals.

Note that gives a solvable group example of the non-implication. See also finitely presented and solvable not implies polycyclic.