Finitely presented not implies Noetherian - Revision history

Vipul: /* Example of the Baumslag-Solitar group */

2013-03-29T01:53:42Z

Example of the Baumslag-Solitar group

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

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

Vipul: /* Proof */

2013-03-29T01:52:13Z

Proof

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	Consider a free group on countably many generators <math>x_n, n \in \mathbb{Z}</math> and now consider an element <math>y</math> that acts on these generators by the rule <math>x_n \mapsto x_{n + 1}</math>. Consider the semidirect product of the free group on the <math>x_n</math>s with a cyclic group generated by <math>y</math>. Clearly, this group is generated by <math>x_0,y</math> and hence is a quotient of <math>\langle a,b \rangle</math> with <math>a \mapsto x_0, b \mapsto y</math>. We see thus that <math>b^nab^{-n} \mapsto x_n</math> and thus the normal closure of <math>\langle a \rangle</math> maps surjectively to the free group on the <math>x_n</math>s. Since the mapping is bijective on a generating set, we see that in fact the normal closure of <math>\langle a \rangle</math> is free on a countable generating set.		Consider a free group on countably many generators <math>x_n, n \in \mathbb{Z}</math> and now consider an element <math>y</math> that acts on these generators by the rule <math>x_n \mapsto x_{n + 1}</math>. Consider the semidirect product of the free group on the <math>x_n</math>s with a cyclic group generated by <math>y</math>. Clearly, this group is generated by <math>x_0,y</math> and hence is a quotient of <math>\langle a,b \rangle</math> with <math>a \mapsto x_0, b \mapsto y</math>. We see thus that <math>b^nab^{-n} \mapsto x_n</math> and thus the normal closure of <math>\langle a \rangle</math> maps surjectively to the free group on the <math>x_n</math>s. Since the mapping is bijective on a generating set, we see that in fact the normal closure of <math>\langle a \rangle</math> is free on a countable generating set.

			===Example of the Baumslag-Solitar group===

			{{further\|[[Particular example::Baumslag-Solitar group:BS(1,2)]]}}

			This group, given by the presentation:

			<math>\langle a,b \mid bab^{-1} = a^2 \rangle</math>

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

Vipul at 17:50, 27 May 2010

2010-05-27T17:50:58Z

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	A [[finitely presented group]] (i.e., a group that admits a finite [[presentation of a group\|presentation]]) need not be a [[Noetherian group]] (also called ''slender group'': a group in which every subgroup is [[finitely generated group\|finitely generated]]).		A [[finitely presented group]] (i.e., a group that admits a finite [[presentation of a group\|presentation]]) need not be a [[Noetherian group]] (also called ''slender group'': a group in which every subgroup is [[finitely generated group\|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==		==Proof==

Vipul: Created page with '{{group property non-implication| stronger = finitely presented group| weaker = Noetherian group}} ==Statement== A finitely presented group (i.e., a group that admits a fin...'

2010-05-27T17:48:11Z

Created page with '{{group property non-implication| stronger = finitely presented group| weaker = Noetherian group}} ==Statement== A finitely presented group (i.e., a group that admits a fin...'

New page

{{group property non-implication|
stronger = finitely presented group|
weaker = Noetherian group}}

==Statement==

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

==Proof==

===Example of the free group of rank two===

{{further|[[Particular example::Free group:F2]]}}

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

Consider a free group on countably many generators <math>x_n, n \in \mathbb{Z}</math> and now consider an element <math>y</math> that acts on these generators by the rule <math>x_n \mapsto x_{n + 1}</math>. Consider the semidirect product of the free group on the <math>x_n</math>s with a cyclic group generated by <math>y</math>. Clearly, this group is generated by <math>x_0,y</math> and hence is a quotient of <math>\langle a,b \rangle</math> with <math>a \mapsto x_0, b \mapsto y</math>. We see thus that <math>b^nab^{-n} \mapsto x_n</math> and thus the normal closure of <math>\langle a \rangle</math> maps surjectively to the free group on the <math>x_n</math>s. Since the mapping is bijective on a generating set, we see that in fact the normal closure of <math>\langle a \rangle</math> is free on a countable generating set.