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.


1 Answer 1


There exists such example (thus also disproving Conjecture 4.1.28 here).

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 $G_\mathbf{R}$.


((1)) (link, MR link) JL Dyer, A nilpotent Lie algebra with nilpotent automorphism group, Bulletin of the American Mathematical Society 76 (1970) 52–56

((2)) (MR link) Favre, Gabriel Une algèbre de Lie caractéristiquement nilpotente de dimension 7. (French) C. R. Acad. Sci. Paris Sér. A-B 274 (1972)

  • $\begingroup$ Note that it's based on the same idea as your solvable example: one choice of group containing the derived subgroup, assuming that the action on [some nontrivial part of] the abelianization is trivial. The only difference is that in the nilpotent world one need to struggle a little further to find the relevant examples; fortunately the cumbersome computational work had already been done (cf references). $\endgroup$
    – YCor
    Oct 17, 2016 at 18:04

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