View source for Finite index not implies local powering-invariant

{{subgroup property non-implication|
stronger = subgroup of finite index|
weaker = local powering-invariant subgroup}}

==Statement==

It is possible to have a [[group]] <math>G</math> and a [[subgroup of finite index]] <math>H</math> of <math>G</math> such that <math>H</math> is not a [[local powering-invariant subgroup]] of <math>G</math>, i.e., there exists an element <math>h \in H</math> and a natural number <math>n</math> such that there is a unique solution <math>u \in G</math> to <math>u^n = h</math>, but <math>u \notin H</math>.

In fact, we can choose <math>H</math> to satisfy any of these additional constraints:

* We can choose <math>H</math> to be a [[characteristic subgroup of finite index]]; in fact, we can choose <math>H</math> to be the [[derived subgroup]].
* We can choose <math>H</math> to be a [[subgroup of index two]].

==Proof==

===Example with subgroup of index two===

Let <math>G</math> be the [[particular example::infinite dihedral group]]:

<math>G := \langle a,x \mid x^2 = e, xax = a^{-1} \rangle</math>.

Here, <math>e</math> denotes the identity element.

Let <math>H</math> be the subgroup <math>\langle a^2,x \rangle</math>.

Then the following are true:

* <math>H</math> is a [[subgroup of index two]] in <math>G</math>.
* <math>H</math> is not local powering-invariant in <math>G</math>. For the element <matH>h = a^2</math> and <math>n = 2</math>, the only solution to <math>u^2 = h</math> for <math>u \in G</math> is <math>u = a</math>, and <math>a \notin H</math>.

===Example with characteristic subgroup of finite index that uses derived subgroup===

Consider the infinite dihedral group, given by the presentation:

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

where <math>e</math> denotes the identity of <math>G</math>. We find that:

<math>[G,G] = \langle a^2 \rangle</math>

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>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.