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I think this question could be a little idiot, however I could not solve this after some hours. I need to find all homomorphism between the additive group $\mathbb{Q}$ and the additive group $\mathbb{Z}$, i.e., $\operatorname{Hom}_{\mathbb{Z}} ( \mathbb{Q}, \mathbb{Z})$. I tried to find a good set of generators for $\mathbb{Q}$, however none of them satisfied a general homomorphism.

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2 Answers 2

up vote 16 down vote accepted

Let $\varphi: \mathbb Q \rightarrow \mathbb Z$ suppose that $\varphi$ is non-trivial. Let $n$ be the smallest positive integer in $\mathrm{im}\; \varphi$. Pick $a/b \in \varphi^{-1}(n)$. Then

$$n=\varphi(a/b)=\varphi(a/2b+a/2b)=\varphi(a/2b)+\varphi(a/2b),$$

so $\varphi(a/2b)=n/2$, but this is impossible because $n$ was the smallest positive integer in the image. Thereby no non-trivial homomorphisms exist.

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$\,\Bbb Q\,$ is a divisible group , so is any of its homomorphic images. But $\,\Bbb Z\,$ is not a divisible group (why?) , so it must be that the image of any group homomorphism $\,\Bbb Q\to\Bbb Z\,$ is the trivial subgroup (which, BTW, is the only finite group which is divisible and the only subgroup of $\,\Bbb Z\,$ that is divisible), and thus $\,\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Z)=0\,$

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"But $\mathbb{Z}$ is not a divisible group, so it must by that the image of any group homomorphism $\mathbb{Q} \rightarrow \mathbb{Z}$ is the trivial subgroup". If you replace $\mathbb{Z}$ by $\mathbb{Q} \oplus \mathbb{Z}$ you can see that there is a gap in your reasoning. You want: more than just that $\mathbb{Z}$ is not divisible, $\mathbb{Z}$ has no nonzero divisible subgroup (a reduced abelian group). Of course the "(why?)" equally well covers this, so it's not a big deal. –  Pete L. Clark Apr 3 '13 at 7:42
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Thanks @PeteL.Clark, I think you're right. The correct, complete argument must be that the only subgroup of the integers that is divisible is the trivial one, though this certainly is contained in the fact that all the subgroups of $\,\Bbb Z\,$ with more than one element are isomorphic with $\,\Bbb Z\,$ ... –  DonAntonio Apr 3 '13 at 9:49

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