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

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* [[Characteristic not implies powering-invariant]]
* [[Characteristic not implies powering-invariant]]
* [[Center is local powering-invariant]]
* [[Center is local powering-invariant]]
* [[Derived subgroup is divisibility-invariant in nilpotent group]]


==Proof==
==Proof==

Revision as of 04:42, 12 February 2013

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

In this example, the generator of the derived subgroup has a unique square root, but this lies outside the derived subgroup (though still in the center). This gives an example where the whole group is a group of nilpotency class two.