Normal subgroup that is invariant under powering such that the quotient group is not invariant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:22:56Z http://mathoverflow.net/feeds/question/121552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121552/normal-subgroup-that-is-invariant-under-powering-such-that-the-quotient-group-is Normal subgroup that is invariant under powering such that the quotient group is not invariant Vipul Naik 2013-02-12T03:21:42Z 2013-02-15T01:45:48Z <p>I want an example of a group $G$, a normal subgroup $H$, and a prime number $p$, such that:</p> <ul> <li>$G$ is powered over $p$, i.e., every element of $G$ has a unique $p^{th}$ root in $G$.</li> <li>$H$ is also powered over $p$, i.e., every element of $H$ has a unique $p^{th}$ root in $H$.</li> <li>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.</li> </ul> <p>While I suspect that an example exists, the example seems hard to construct, because of the following constraints I worked out for any example:</p> <ul> <li>$H$ must be infinite and have infinite index in $G$ (i.e., neither $H$ nor $G/H$ can be finite).</li> <li>$H$ 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.</li> <li>$H$ cannot have a complement (i.e., be part of a semidirect product) in $G$.</li> </ul> <p>The proofs of all these assertions are straightforward, but I'll be happy to provide proofs if they are unclear to readers.</p> <p>If you find a proof that no such example exists, that would be great to have too.</p> http://mathoverflow.net/questions/121552/normal-subgroup-that-is-invariant-under-powering-such-that-the-quotient-group-is/121647#121647 Answer by Denis Osin for Normal subgroup that is invariant under powering such that the quotient group is not invariant Denis Osin 2013-02-12T20:39:47Z 2013-02-15T01:45:48Z <p>This is a corrected answer. I apologize for not posting a complete proof here.</p> <p>Recall that a group $G$ is called <em>divisible</em> 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 $D$. </p> <p>The following theorem answers the question. </p> <p><strong>Theorem.</strong> <em>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 $n\in \mathbb N$.</em></p> <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 <a href="http://arxiv.org/abs/math/0411039" rel="nofollow">http://arxiv.org/abs/math/0411039</a>. </p> <p>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). </p>