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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) 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).

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).

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) 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.

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) 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).

This is a corrected answer. I apologize for not posting a complete proof here.

Recall that a group $G$ is called divisible (respectively, uniquely divisible) if for every $g\in G$ and $n\in \mathbb N$, there is $x\in G$ (respectively, there is unique $x\in G$) satisfying $x^n=g$. Recall also that there exist countable (and even finitely generated) torsion free groups where all nontrivial elements are conjugate; we fix one such a group and denote it by $C$. Obviously $C$ is divisible, but not uniquely divisible; moreover, for every prime $p$ and every $g\in C\setminus\{ 1\}$, $g$ has infinitely many $p$th roots.

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 $C$. In particular, there are elements $g\in G/H$ that have infinitely many $p$th roots for every prime $p$.

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.

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) 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.