I want an example of a nilpotent group $G$, a characteristic subgroup $H$, and a prime number $p$ such that:


* $G$ is $p$-powered, i.e., every element of $G$ has a unique $p^{th}$ root in $G$.
* $H$ is not $p$-powered. In this case, it would mean that there exists an element of $H$ that does not have any $p^{th}$ root in $H$.


Note that there do not exist any abelian examples, because in an abelian $p$-powered group, the $p^{th}$ root map is an automorphism and hence preserves every characteristic subgroup.


The intuitive reasoning for why there shouldn't be any examples: even though the $p^{th}$ root map is not an automorphism, it seems possible generally to find automorphisms that behave a lot like this map to a sufficient extent that characteristic subgroups must be preserved under taking $p^{th}$ roots. 


The intuitive reasoning for why there may well be examples: nilpotent groups can in principle rigidify a lot of powering structure.


I don't know which side to believe.


PS: There are certainly characteristic subgroups of non-nilpotent groups with this property. For instance, let $G = GA^+(1,\mathbb{R})$ (the identity component of the affine group of degree one over the reals, i.e., maps of the form $x \mapsto ax + b, a > 0, a,b \in \mathbb{R}$), and let $H$ be the subgroup comprising those elements where $a$ is rational. $H$ is characteristic in $G$, $G$ is powered over all primes, and $H$ is not powered over any prime.



Characteristic subgroup of nilpotent group that is not invariant under powering
There exists such example (thus also disproving Conjecture 4.1.28 [here](http://files.vipulnaik.com/thesis/thesis.pdf)).


Indeed there exists a nonzero nilpotent Lie algebra $\mathfrak{g}$ with rational coefficients whose automorphism group $A$ is unipotent (references (1,2) below). Therefore $\mathfrak{g}$ admits an $A$-invariant $\mathbf{Q}$-defined proper subalgebra $\mathfrak{l}$ such that the induced $A$-action on $\mathfrak{g}/\mathfrak{h}$ is trivial. 


Let $G$ is the corresponding unipotent $\mathbf{Q}$-group ($G\_K$ can be defined by endowing $\mathfrak{g}\_K$ with the group law defined by the Baker-Campbell-Hausdorff formula) and $L\_\mathbf{Q}$ the corresponding subgroup, so that $\mathrm{Aut}(G\_\mathbf{Q})=A\_{\mathbf{Q}}$. Then $L\_\mathbf{Q}$ is a characteristic subgroup of $G\_\mathbf{Q}$ and the action of $\mathrm{Aut}(G\_\mathbf{Q})$ on $G\_\mathbf{Q}/L\_\mathbf{Q}$ is trivial. It follows that every subgroup of $G\_\mathbf{Q}$ containing $L\_\mathbf{Q}$ is also a characteristic subgroup. So, if we define $H\subset G\_\mathbf{Q}$ as the inverse image of an infinite cyclic subgroup of $G\_\mathbf{Q}/L\_\mathbf{Q}$, then $H$ is not $p$-divisible (hence not $p$-powered) for any prime $p$, while $G\_\mathbf{Q}$ is $p$-powered for all primes $p$. 


For those who rather like Lie groups: if we perform the same construction in the real case, we get $H\subset G\_\mathbf{R}$, where $G\_\mathbf{R}$ is a simply connected nilpotent Lie group, $H$ is a closed proper subgroup closed under the whole group of continuous group automorphisms of $G\_\mathbf{R}$.




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**References**


((1)) ([link](https://projecteuclid.org/euclid.bams/1183531386), [MR link](http://www.ams.org/mathscinet-getitem?mr=249544)) JL Dyer, A nilpotent Lie algebra with nilpotent automorphism group, Bulletin of the
American Mathematical Society 76 (1970) 52–56


((2)) [(MR link)](http://www.ams.org/mathscinet-getitem?mr=294425) Favre, Gabriel
Une algèbre de Lie caractéristiquement nilpotente de dimension 7. (French) 
C. R. Acad. Sci. Paris Sér. A-B 274 (1972)


