Does there exist any surjective group homomorphism from $(\mathbb R^* , .)$ onto $(\mathbb Q^* , .)$? 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)? 
 A: 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. 
A: $f(-1)^2=f(1)$ implies that $f(-1)=1$ or $f(-1)=-1$. Suppose $f(-1)=1$. Let $x>0$, $f(x)=f(\sqrt x)^2$. Implies that $f(x)$ is a square, $f(-x)=f(-1)f(x)>0$. Impossible. Since $f$ is surjective.
Suppose $f(-1)=-1, f(-x)=-f(\sqrt x)^2$ and $f(x)=f(\sqrt x)^2$ this implies that $f$ is not surjective since $1/2$ is not in the image: it is positive but not the square of a rational.
A: Suppose that $f:\Bbb{R}^*\to\Bbb{Q}^*$ is a such homomorphism. Then we have some $x$ such that $f(x)=2$. Now take a cube root $\sqrt[3]{x}$ of $x$, which always exists in $\mathbb{R}^*$. Then $(f(\sqrt[3]{x}))^3 = f(x) = 2$, i.e. $f(\sqrt[3]{x})$ is a cube root of $2$. But $2$ has no cube root in $\mathbb{Q}^*$, so this is a contradiction.
