Characterize all subgroups of $\displaystyle \mathbb{Q}^{\times}.$
My guess would be that $\displaystyle ~ \mathbb{Q}^{\times}_{>0} = \displaystyle\bigoplus_{p ~ prime} \langle p \rangle ~$. Clearly each $\displaystyle \langle p \rangle \simeq \mathbb{Z}$
Then $\displaystyle \mathbb{Q}^{\times} \simeq G \oplus \mathbb{Q}^{\times}_{>0}$ as we can define the trivial isomorphism $\displaystyle \theta$ : $\displaystyle (g,q) \rightarrow gq$
Just the first step needs some justification...
well, every $\displaystyle r \in \mathbb{Q}^{\times}$ can be written uniquely as $\displaystyle r=\pm p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k},$ where $\displaystyle p_j$ is the j-th prime number and $\displaystyle n_j \in \mathbb{Z}.$ now define $\displaystyle f(r)=(\text{sgn}(r),n_1,n_2, \cdots, n_k, 0, 0, \cdots ).$
EDIT: Aw man, ninja'd ><
Original post:
As a consequence of uniqueness of prime factorization, every nonzero rational number can be expressed uniquely as $\displaystyle q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots p_s^{r_s}$ where $\displaystyle r_i \in \mathbb{Z}^\times$.
Then given a particular enumeration of the primes $\displaystyle P=\{p_1, p_2, \ldots\}$, there are only a finite number of prime divisors, so with respect to our enumeration there is a unique such representation with $\displaystyle q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots =(-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i}$ where $\displaystyle r_i \in \mathbb{Z}$ such that only a finite number of $\displaystyle r_i $ are nonzero This would suggest an isomorphism $\displaystyle Q^\times \rightarrow G\times\displaystyle\bigoplus_{i=1}^\infty \mathbb{Z}$ given by $\displaystyle (-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i} \mapsto ((-1)^{r_0}, r_1, r_2, \ldots)$.
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 $\displaystyle N$ be a subgroup of $\displaystyle \mathbb{Q}^{\times}.$ then either $\displaystyle \{1,-1 \}=G \subseteq N$ or $\displaystyle N \cap G = \{1 \}.$ if $\displaystyle G \subseteq N,$ then $\displaystyle N/G$ would be a subgroup of the free abelian group $\displaystyle \mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ we know that any
(non-trivial} subgroup of a free abelian group is free. so $\displaystyle N$ is isomorphic to $\displaystyle G \oplus K,$ where $\displaystyle K$ is the direct sum of finitely or infinitely (countable) many copies of $\displaystyle \mathbb{Z}.$ if $\displaystyle N \cap G = \{1 \},$ then
$\displaystyle NG/G \cong N/(N \cap G) \cong N.$ thus $\displaystyle NG/G \cong N$ is a subgroup of the free abelian group $\displaystyle \mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ so $\displaystyle N$ is just a direct sum of finitely or infinitely (countable) many of copies of $\displaystyle \mathbb{Z}.$
briefly subgroups of $\displaystyle \mathbb{Q}^{\times}$ are in one of these forms: $\displaystyle \{1 \}, \ \{1,-1 \}, \ K, \ \{1,-1 \} \oplus K,$ where $\displaystyle K$ is any free abelian group of at most countable rank.