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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 $\mathbb{Z} \to \mathbb{Q}^{\times}$?

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    It's probably not boring if you have to ask! That is, before you are aware of such a fact. And even once the fact is known, you could ask (a student) further questions: is there any other single element of $\mathbb{Z}$ that induces a homomorphism? How many are there in total? How do you know? More generally, when we have a homomorphism $f: K \rightarrow G$, for what sorts of groups $K$ will a single element determine the homomorphism? Why? etc. [Would you say the reverse question for $\mathbb{Q}^{\times} \rightarrow \mathbb{Z}$ is boring too? For anyone who thinks $p$-adically...]2012-10-29

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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 $f(n)=f(1)^n$).

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    @OrestXherija: $e_p$ is the power of the prime $p$ in a given rational, positive if it is in the numerator and negative if it is in the denominator. The example of $\frac {20}{363}$ shows this.2012-10-28