Krdytkyu

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)?

Suppose that $f:Bbb{R}^*toBbb{Q}^*$ 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.

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 $ain G$, 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.

$f(-1)^2=f(1)$ 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.