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 $y^p = x$.

Question:

- 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*.