Let p be a prime number. Call a group G uniquely p-divisible if for every x in G there is a unique y in G such that $y^p = x$.


  1. Must a characteristic subgroup of a uniquely p-divisible group also be uniquely p-divisible? In symbols, if H is a characteristic subgroup of a uniquely p-divisible group G, must H also be uniquely p-divisible?
  2. Is the statement true if we impose the additional condition that G is a nilpotent group? The condition of being nilpotent is often a pretty strong restriction on the existence of various kinds of roots, so I think this is much more plausible.

What I know:

A. The statement is true if the big group G is abelian. This is because multiplication and division by p become automorphisms and hence must preserve any characteristic subgroup.

B. The statement is true if G is finite. In fact, if G is finite, this is equivalent to saying that the order of G is relatively prime to p, and hence all subgroups are uniquely p-divisible.

C. For infinite groups, we can have non-characteristic subgroups that violate the condition. For instance, the group of integers in the group of rational numbers. The big group is uniquely p-divisible for all p, but the group of integers is not uniquely p-divisible for any p.


The first example of a finitely generated divisible group was constructed by Guba in Guba, V. S. A finitely generated complete group. Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986), no. 5, 883–924. Similarly one constructs a $p$-divisible group. Start with the free group $F=F(a,b)$, enumerate all non-identity elements of $F$: $w_0,w_1...,$. Then let $G_0=F$, and at every step $n \ge 0$, if the word $w_n$ is a $p$-th power in $G_n$, do nothing, otherwise impose one additional relation $w_n=U_n^p$, where $U_n$ is "long and complicated". The new group is $G_{n+1}$. It is easy to see that the limit group $G_\infty$ is $p$-divisible. It is possible to prove also that it is going to be uniquely $p$-divisible, provided $U_n$ satisfy some small cancelation condition. Now, let $w_0=[a,b]$ be the commutator (it does not matter how you enumerate the words in $F$), and $U_0$ be not in the derived subgroup (the choice of $U_0$ is almost arbitrary). Then the derived subgroup of $G_\infty$ will not be $p$-divisible. The derived subgroup is of course characteristic. This answers your first question. For nilpotent groups the answer may be different indeed.


I think the approach suggested by Mark works, but one has to be a bit more careful. The problem is that, in Mark's notation, if $U_0$ is not in the derived subgroup of $F$, it does not follow automatically that the image of $U_0$ will not be in the derived subgroup of $G_\infty $. To ensure this property one has to be more careful with the choice of other $U_i$'s.

The best illustration of this problem is the following observation: if $G$ is finitely generated and divisible, then $[G,G]=G$ and in particular $[G,G]$ is divisible. Indeed, $G/[G,G]$ is a finitely generated divisible abelian group, hence it is trivial, i.e. $[G,G]=G$. Nevertheless, this approach should work after some modifications, which are obvious to experts (e.g., to Mark).

Alternatively, one can find examples of uniquely p-divisible groups whose derived subgroups are not p-divisible in the old paper [G. Baumslag, Some aspects of groups with unique roots, Acta Math. 104 1960 217–303]. These are so-called free $D_\omega $-groups.

If I am not mistaken, in the same paper Baumslag proved that every term of the lower central series, upper central series, or derived series of a uniquely divisible nilpotent group is again (uniquely) divisible. However I do not see why it should be so for other verbal subgroups.

  • $\begingroup$ Thanks for the references and issues you raised! As for your example of $G^p$, if I understand your notation correctly, then G^p = G when G is uniquely p-divisible, so that couldn't generate a counterexample. You probably had something slightly different in mind. $\endgroup$ – Vipul Naik Mar 1 '11 at 22:03
  • $\begingroup$ Yes, I agree. I corrected the answer. $\endgroup$ – Denis Osin Mar 2 '11 at 1:05

The example of $GA^+(1,\mathbb{R})$ here works as a counterexample:



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