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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)?

Context

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.

1 Answers 1

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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...