# Finite index not implies local powering-invariant

## 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.