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 $f:\Bbb{R}^*\to\Bbb{Q}^*$ 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.


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.

  • $\begingroup$ the only divisible subgroup of $(\mathbb Q^* , .)$ is trivial ... I can feel it intuitively but how do you prove that rigorously ? could you please elaborate ? $\endgroup$ – user228168 Aug 20 '16 at 3:30
  • $\begingroup$ The point is that you cannot take arbitrary roots of a rational number (in the rationals). Suppose $a$ would allow a solution, in the rationals, to $X^n =a$ for each $a$. We show $a=1$ We can focus on $a$ positive, as a negative number has no squareroot. Thus $a = p_1^{v_1} \dots p_k^{v_k}$ with distinct primes $p_i$ and integers $v_i$. Now for $X^n = a$ to have a solution we need $n \mid v_i$ for each $i$. For each fixed $a$, this thus cannot have a solution for all $n$ unless all the $v_i$ are $0$ that is $a=1$. $\endgroup$ – quid Aug 20 '16 at 8:51
  • $\begingroup$ Differently, the multiplicative group of positve rationals is isomorphic to the additive group $\mathbb{Z}^{(\mathbb{N})}$. The latter is not divisible. $\endgroup$ – quid Aug 20 '16 at 8:51

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

  • $\begingroup$ This answer could use serious rewriting, because as it is it's really not clear. If someone read this and didn't already know the proof I doubt they'd understand what's going on. $\endgroup$ – Najib Idrissi Aug 19 '16 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy