TITLE: Does there exist any surjective group homomorphism from $(\mathbb R^* , .)$ onto $(\mathbb Q^* , .)$?
QUESTION [7 upvotes]: Does there exist any surjective group homomorphism from $(\mathbb R^* , .)$ (the multiplicative group of non-zero real numbers) onto $(\mathbb Q^* , .)$ (the multiplicative group of non-zero rational numbers)?

REPLY [3 votes]: Let me add a more abstract view to the concrete arguments already given. 
Recall that a group $(G,\cdot)$ is called divisible if for each $a\in G$, positive integer $n$, the equation $X^n = a$ has a solution. 
(The terminology makes more sense in additive notation where it means $nX=a$ has a solution; that is, one  can 'divide the $a$ into $n$ equal parts.')
Now observe:

the homomorphic image of a divisible group is divisible.  
the only divisible subgroup of $(\mathbb{Q}^{\ast}, \cdot)$ is the trivial  one.
the (sub)group $(\mathbb{R}_+^{\ast}, \cdot)$ is divisible.

Thus, the image of $(\mathbb{R}_+^{\ast}, \cdot)$ under every homorphism $\varphi$ from $(\mathbb{R}^{\ast}, \cdot)$ to $(\mathbb{Q}^{\ast}, \cdot)$ must be trivial.
Now, for $x$ negative we have that $x^2$ is positive and thus $\varphi(x)^2=\varphi(x^2)= 1$. Thus, $\varphi(x)$ is $\pm 1$. (Also see another answer for this.)
Assume there are two negative number $x,y$ such that $\varphi(x) \neq \varphi (y)$, then $1= \varphi(xy) = \varphi(x)\varphi(y) = -1$, a contradiction. 
Thus either all negative numbers have image $1$ or all negative number have image $-1$. 
It follows that the only two homomorphism there could be are $x \mapsto 1$ and $x \mapsto \operatorname{sign}{(x)}$. 
Both are indeed homomorphisms, yet neither is surjective.