here's an idea: let considered as a multiplicative group. show that and are isomorphic.
EDIT: Aw man, ninja'd ><
As a consequence of uniqueness of prime factorization, every nonzero rational number can be expressed uniquely as where .
Then given a particular enumeration of the primes , there are only a finite number of prime divisors, so with respect to our enumeration there is a unique such representation with where such that only a finite number of are nonzero This would suggest an isomorphism given by .
Uniqueness of factorization assures this is a well defined and 1-1 function. The finite number of prime factors assures this indeed maps into and onto the direct sum. Finally the morphism property follows from the properties of exponents.
Hand wavy but conveys the ideas I think.
Either the subgroups are finitely generated or not. If they are finitely generated, then the fundamental theorem of finitely generated abelian groups tells us that up to isomorphisms the subgroups are the direct product of copies of the integers with the 2 element group. If a subgroup is not finitely generated, then intuition says we should expect an isomorphic copy of the group, or an isomorphic copy of the infinite direct sum of integers.
let be a subgroup of then either or if then would be a subgroup of the free abelian group we know that any
(non-trivial} subgroup of a free abelian group is free. so is isomorphic to where is the direct sum of finitely or infinitely (countable) many copies of if then
thus is a subgroup of the free abelian group so is just a direct sum of finitely or infinitely (countable) many of copies of
briefly subgroups of are in one of these forms: where is any free abelian group of at most countable rank.