Derived subgroup not is local powering-invariant

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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)
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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 u \in G satisfying u^n = h, and despite this, u \notin H.

We can choose G to be a metacyclic group. We could also choose G to be a finitely generated nilpotent group, and in fact an example of a finitely generated group of nilpotency class two.

Related facts


Example of the infinite dihedral group (metacyclic example)

Further information: infinite dihedral group

Consider the infinite dihedral group, given by the presentation:

G := \langle a,x \mid xax^{-1} = a^{-1}, x^2 = e \rangle

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

[G,G] = \langle a^2 \rangle

is an infinite cyclic group.

Now consider the element h = a^2. Let n = 2. We note that all elements outside \langle a \rangle have order two, hence any element u with u^2 = h must be inside \langle a \rangle. The only possibility is thus u = 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 (finitely generated group of nilpotency class two)

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.