# 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.

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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) 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).

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Denis, I read the question as asking for groups in which pth roots always exist. –  HJRW Feb 12 '13 at 20:47
HW's right, I want an example where $p^{th}$ roots always exist and are unique for both $G$ and $H$. –  Vipul Naik Feb 12 '13 at 20:56
You are right, I did not notice that. Then it is harder. I'll edit the answer. –  Denis Osin Feb 12 '13 at 21:01
Yes, this is exactly what I want. If I'm reading your paper correctly, this doesn't follow immediately from the result you reference, but probably could with some work. Thanks a lot! I'll read through the rest of your paper to see if it addresses some related questions. –  Vipul Naik Feb 13 '13 at 0:32
Yes, it does not follow directly. You need to essentially modify some arguments in the proof of Theorem 1.5, but all necessary tools are there. –  Denis Osin Feb 13 '13 at 3:24