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Let $f : (\mathbb{Q},+) \longrightarrow (\mathbb{Q},+)$ be a non-zero homomorphism.

Can we conclude that $f$ is bijective (or, if that fails, that $f$ is injective or surjective)?


The additive group of integers has non-surjective nonzero endomorphisms, such as $n\mapsto 2n$. However, the same formula gives a bijective endomorphism when applied to rationals.

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Have you had any thoughts about the problem? – Thomas Aug 6 '12 at 17:02
What do you think? Hint: start with the definition of "homomorphism" and what you know about $\mathbb{Q}$. – Code-Guru Aug 6 '12 at 17:04
Oh, and it can't be option (c) if there is only one true statement. If (c) was true, then (a) and (b) would also be true. – Thomas Aug 6 '12 at 17:18
up vote 8 down vote accepted

Suppose $\,f\,$ is such a homomorphism and try to work out how knowing $\,f(1)\,$ can help you out, say: $$\forall\,n\in\Bbb Z\,\,,\,\,f(n)=f(n\cdot 1)=nf(1)$$ $$f(1)=f\left(\frac{n}{n}\right)=nf\left(\frac{1}{n}\right).....etc.$$

So you know the values of $\,f\,$ on the integers and then on the rationals of the form $\,1/n\,$ and thus...

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dilates eyes...... – David Wheeler Aug 6 '12 at 17:44

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