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 functionsView subgroup property dissatisfactions for subgroup-defining functions
Statement
It is possible to have a group such that the derived subgroup is not a local powering-invariant subgroup of . Specifically, it is possible that there exists an element and a natural number such that there exists a unique element satisfying but .
Related facts
Proof
Example of the infinite dihedral group
Further information: infinite dihedral group
Consider the infinite dihedral group, given by the presentation:
where denotes the identity of . We find that:
is an infinite cyclic group.
Now consider the element . Let . We note that all elements outside have order two, hence any element with must be inside . The only possibility is thus , which is outside . Thus, the element has a unique square root in , but this is not in , completing the proof.
Example of a central product
Further information: central product of UT(3,Z) and Z identifying center with 2Z