Derived subgroup not is local powering-invariant

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 $h\in [G,G]$ and a natural number $n$ such that there exists a unique element $w\in G$ satisfying w^n = h</math> but $w\notin H$.

Proof

Example of the infinite dihedral group

Further information: infinite dihedral group

Consider the infinite dihedral group, given by the presentation:

$G:=⟨a,x\mid xa{x}^{-1}=a^{-1},x^2=e⟩$

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

$[G,G]=⟨a^2⟩$

is an infinite cyclic group.

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

Example of a central product

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