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 $y^p = x$.
- 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?
- 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.