# Finite index not implies local powering-invariant

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of finite index) neednotsatisfy the second subgroup property (i.e., local powering-invariant subgroup)

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## Contents

## Statement

It is possible to have a group and a subgroup of finite index of such that is not a local powering-invariant subgroup of , i.e., there exists an element and a natural number such that there is a unique solution to , but .

In fact, we can choose to satisfy any of these additional constraints:

- We can choose to be a characteristic subgroup of finite index; in fact, we can choose to be the derived subgroup.
- We can choose to be a subgroup of index two.

## Proof

### Example with subgroup of index two

Let be the infinite dihedral group:

.

Here, denotes the identity element.

Let be the subgroup .

Then the following are true:

- is a subgroup of index two in .
- is not local powering-invariant in . For the element and , the only solution to for is , and .

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

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.