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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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Asking about homomorphisms $f : \mathbb{Z} \to G$ for any group $G$ is boring: any such $f$ is freely and uniquely determined by $f (1)$. –  Zhen Lin Oct 28 '12 at 23:45
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...] –  Benjamin Dickman Oct 29 '12 at 0:07

1 Answer 1

up vote 10 down vote accepted

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

Each rational $q$ can be uniquely represented in the form $$q=\prod_{p \text{ prime}} p^{e_p}$$

for some integers $e_p$. 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 $g(\pm \prod p^{e_p})= \sum e_p n_p$.

A homomorphism $f\colon \mathbb Z\to\mathbb Q^\times$ is determined by selecting $f(1)$ arbitrarily.

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Thanks a lot! For the second part, did you meant $\mathbb{Q}^x$ or $\mathbb{Q}$? Also, I am not sure what your $e_p$ denotes. –  user44069 Oct 28 '12 at 22:41
@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. –  Ross Millikan Oct 28 '12 at 23:38

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