Non-trivial homomorphism between multiplicative group of rationals and integers

Let $\mathbb{Q}^{\times}$ be the multiplicative group of non-zero rationals. Is there a non-trivial homomorphism $\mathbb{Q}^{\times} \to \mathbb{Z}$? In the same spirit, is there a homomorphism $\mathbb{Z} \to \mathbb{Q}^{\times}$?

Solution 1:

For every prime $p$, select an integer $n_p$.

Each rational $q$ can be uniquely represented in the form $$q=\pm\prod_{p \text{ prime}} p^{e_p}$$ for some integers $e_p$, almost all of which are $0$. For example, $${20\over 363} = {2^2\cdot 5\over 3\cdot 11^2} = 2^2\cdot 3^{-1}\cdot5^1\cdot7^0\cdot11^{-2}\cdot13^0\cdot17^0\cdots.$$ Here $e_2 = 2$, $e_3 = -1 $, and so forth.

Then map $g\colon\mathbb Q^\times\to\mathbb Z$ by letting $g(\pm \prod p^{e_p})= \sum e_p n_p$. This is a homomorphism (and one can show that all homomorphisms $\Bbb Q^\times \to\Bbb Z$ can be obtained this way; that is, given a homomorphism $f$, the homomorphism $g$ obtained by picking $n_p:=f(p)$ equals $f$)

A homomorphism $f\colon \mathbb Z\to\mathbb Q^\times$ is determined by selecting $f(1)\in\Bbb Q^\times$ arbitrarily (and letting $f(n)=f(1)^n$).