Derived subgroup not is local powering-invariant: Difference between revisions

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It is possible to have a [[group]] <math>G</math> such that the [[derived subgroup]] <math>[G,G]</math> is ''not'' a [[local powering-invariant subgroup]] of <math>G</math>. Specifically, it is possible that there exists an element <math>h \in [G,G]</math> and a natural number <math>n</math> such that there exists a unique element <math>w \in G</math> satisfying <math>w^n = h</math> but <math>w \notin H</math>.
It is possible to have a [[group]] <math>G</math> such that the [[derived subgroup]] <math>[G,G]</math> is ''not'' a [[local powering-invariant subgroup]] of <math>G</math>. Specifically, it is possible that there exists an element <math>h \in [G,G]</math> and a natural number <math>n</math> such that there exists a unique element <math>w \in G</math> satisfying <math>w^n = h</math> but <math>w \notin H</math>.
==Related facts==
* [[Characteristic not implies powering-invariant]]
* [[Center is local powering-invariant]]


==Proof==
==Proof==

Revision as of 04:21, 7 June 2012

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) does not always satisfy a particular subgroup property (i.e., local powering-invariant subgroup)
View subgroup property satisfactions for subgroup-defining functions

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View subgroup property dissatisfactions for subgroup-defining functions

Statement

It is possible to have a group G such that the derived subgroup [G,G] is not a local powering-invariant subgroup of G. Specifically, it is possible that there exists an element h[G,G] and a natural number n such that there exists a unique element wG satisfying wn=h but wH.

Related facts

Proof

Example of the infinite dihedral group

Further information: infinite dihedral group

Consider the infinite dihedral group, given by the presentation:

G:=a,xxax1=a1,x2=e

where e denotes the identity of G. We find that:

[G,G]=a2

is an infinite cyclic group.

Now consider the element h=a2. Let n=2. We note that all elements outside a have order two, hence any element w with w2=h must be inside a. The only possibility is thus w=a, which is outside H. Thus, the element h=a2 has a unique square root in G, but this is not in H, completing the proof.

Example of a central product

Further information: central product of UT(3,Z) and Z identifying center with 2Z