# 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) need not satisfy the second subgroup property (i.e., local powering-invariant subgroup)
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## Statement

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

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

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

## Proof

### Example with subgroup of index two

Let $G$ be the infinite dihedral group: $G := \langle a,x \mid x^2 = e, xax = a^{-1} \rangle$.

Here, $e$ denotes the identity element.

Let $H$ be the subgroup $\langle a^2,x \rangle$.

Then the following are true:

• $H$ is a subgroup of index two in $G$.
• $H$ is not local powering-invariant in $G$. For the element $h = a^2$ and $n = 2$, the only solution to $u^2 = h$ for $u \in G$ is $u = a$, and $a \notin H$.

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

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.