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

Latest revision as of 15:31, 24 June 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 $h \in [G, G]$ 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

Proof

Example of the infinite dihedral group (metacyclic example)

Further information: infinite dihedral group

Consider the infinite dihedral group, given by the presentation:

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

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 $u$ with $u^{2} = h$ must be inside $⟨ a ⟩$ . 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.

@@ Line 1: / Line 1: @@
 {{sdf subgroup property dissatisfaction|
 sdf = derived subgroup|
-subgroup property = local powering-invariant subgroup}}
+property = local powering-invariant subgroup}}
 ==Statement==
-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 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>u \in G</math> satisfying <math>u^n = h</math>, and despite this, <math>u \notin H</math>.
+We can choose <math>G</math> to be a [[metacyclic group]]. We could also choose <math>G</math> to be a [[finitely generated nilpotent group]], and in fact an example of a finitely generated [[group of nilpotency class two]].
+==Related facts==
+* [[Characteristic not implies powering-invariant]]
+* [[Center is local powering-invariant]]
+* [[Derived subgroup is divisibility-invariant in nilpotent group]]
 ==Proof==
-===Example of the infinite dihedral group===
+===Example of the infinite dihedral group (metacyclic example)===
 {{further|[[particular example::infinite dihedral group]]}}
@@ Line 23: / Line 31: @@
 is an [[infinite cyclic group]].
-Now consider the element <math>h = a^2</math>. Let <math>n = 2</math>. We note that all elements outside <math>\langle a \rangle</math> have order two, hence any element <math>w</matH> with <math>w^2 = h</math> must be inside <math>\langle a \rangle</math>. The only possibility is thus <math>w = a</math>, which is outside <math>H</math>. Thus, the element <matH>h = a^2</math> has a unique square root in <math>G</math>, but this is not in <math>H</math>, completing the proof.
+Now consider the element <math>h = a^2</math>. Let <math>n = 2</math>. We note that all elements outside <math>\langle a \rangle</math> have order two, hence any element <math>u</math> with <math>u^2 = h</math> must be inside <math>\langle a \rangle</math>. The only possibility is thus <math>u = a</math>, which is outside <math>H</math>. Thus, the element <matH>h = a^2</math> has a unique square root in <math>G</math>, but this is not in <math>H</math>, completing the proof.
+===Example of a central product (finitely generated group of nilpotency class two)===
-===Example of a central product===
+{{further|[[particular example::central product of UT(3,Z) and Z identifying center with 2Z]]}}
-{{further|[[particular example::semidirect 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]].