Normal subgroup that is invariant under powering such that the quotient group is not invariant I want an example of a group $G$, a normal subgroup $H$, and a prime number $p$, such that:


*

*$G$ is powered over $p$, i.e., every element of $G$ has a unique $p^{th}$ root in $G$.

*$H$ is also powered over $p$, i.e., every element of $H$ has a unique $p^{th}$ root in $H$.

*The quotient group $G/H$ is not powered over $p$. Since the above conditions already guarantee the existence of $p^{th}$ roots, what I want should fail is the uniqueness condition.


While I suspect that an example exists, the example seems hard to construct, because of the following constraints I worked out for any example:


*

*$H$ must be infinite and have infinite index in $G$ (i.e., neither $H$ nor $G/H$ can be finite).

*$H$ cannot be contained in the hypercenter of $G$ (the hypercenter is the subgroup at which the upper central series stabilizes). This rules out any example involving $G$ abelian or nilpotent.

*$H$ cannot have a complement (i.e., be part of a semidirect product) in $G$.


The proofs of all these assertions are straightforward, but I'll be happy to provide proofs if they are unclear to readers.
If you find a proof that no such example exists, that would be great to have too.
 A: This is a corrected answer. I apologize for not posting a complete proof here.
Recall that a group $G$ is called divisible if for every $g\in G$ and $n\in \mathbb N$, there is $x\in G$ satisfying $x^n=g$. Recall also that there exist countable (and even finitely generated) torsion free divisible groups where every element has infinitely many $n$th roots for every $n$. We fix one such a group and denote it by $D$. 
The following theorem answers the question. 
Theorem.
There exists a countable uniquely divisible group $G$ and a divisible normal subgroup $H\le G$ such that $G/H$ contains $D$. In particular, there are elements $g\in G/H$ that have infinitely many $n$th roots for every $n\in \mathbb N$.
Unfortunately, I do not know any easy proof. The only proof I know would take few pages. The main idea is to use a modification of the construction from the proof of Theorem 1.5 of my paper http://arxiv.org/abs/math/0411039. 
These groups $G$ and $H$ are very far from being finite or nilpotent; they will contain non-abelian free subgroups (this is unavoidable in my construction). 
