Normal subgroup that is invariant under powering such that the quotient group is not invariant - MathOverflow

Normal subgroup that is invariant under powering such that the quotient group is not invariant

2013-02-12T03:21:42Z

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.

Answer by Denis Osin for Normal subgroup that is invariant under powering such that the quotient group is not invariant

2013-02-12T20:39:47Z

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).