Onto group homomorphism How many onto group homomorphism is possible from $(\mathbb{Q},+)$ onto $(\mathbb{Q}\setminus \lbrace 0\rbrace,\cdot)$?
 A: The group $(\mathbb{Q}, +)$ is divisible, and so is every homomorphic image of it. But $(\mathbb{Q} - \{0 \}, \cdot)$ is not divisible: $\sqrt{2}$ is irrational. Hence there can be no surjective homomorphism between the two groups given. 
Note that any rational number that is not $1$ has an irrational $n$th root for some $n$, so with similar reasoning you can prove a stronger result: any homomorphism $(\mathbb{Q}, +) \rightarrow (\mathbb{Q} - \{0\}, \cdot)$ must be trivial.
A: None.
Let $\phi\colon \mathbb Q\to\mathbb Q^\times$ be a homomorphism.
Then only finitely many primes occur as factor (in positive or negative power) in $f(1)$.
On the other hand, for $x\in \mathbb Q$ with $x>0$, we have $\underbrace{x+\ldots+x}_n=\underbrace{1+\ldots +1}_m$ for some $n,m\in\mathbb N$, hence $f(x)^n=f(1)^m$. We conclude that $f(x)$ does not involve any primes beyond those occuring in $f(1)$. For $x<0$, the situation is the same as $f(x)=\frac1{f(-x)}$.
Since there are infinitely primes, $\phi$ is not onto.
A: I guess I'll post my comment as an answer: not only are there no onto homomorphisms, but every homomorphism $\phi: (\mathbb{Q}, +) \to (\mathbb{Q}^\times, \cdot)$ must be trivial (i.e., $\phi$ must send every element to $1$). For if $a \in \mathbb{Q}$ and $\phi(a) = (-1)^r \cdot \prod_{i=1}^k p_i^{e_i}$, then putting $b = a/k$ for any given $k > 0$, we see $\phi(b)^k = \phi(a)$, so that $k$ must divide every $e_i$; this is impossible to have for all $k$ unless all the $e_i$ are zero. Replacing $k$ by $2 k$, we similarly conclude $r$ must be even. Hence $\phi(a) = 1$ for any $a$. 
