# A homomorphism from $(\mathbb{Q},+)\rightarrow (\mathbb{Q},+)$

I claim that any group homomorphism will be of the form $f(x)=qx$ for some $q\in\mathbb{Q}$, now as $f$ is non zero so $q\neq 0$ now clearly kernel of $f$ will be $\{0\}$ only hence $f$ is injective, now let $f(1)=p\neq 0$ then for any $y\in\mathbb{Q}$ and $f(y/p)=y$ so $f$ is surjective too, hence bijective. am I right?

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You'd have to prove that $f(y/p)=y$ from $f(1)=p$. Not hard, but not one step. – Thomas Andrews Dec 18 '12 at 19:04
Though I knew the proof but thank you very much for the response :) – Taxi Driver Dec 18 '12 at 19:16

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Yes, it just wasn't clear from the post if you thought it was "obvious" that if $f(1)=p$ then $f(y/p)=y$ for all $y$, or if you just were skipping the proof because you were done with that part. – Thomas Andrews Dec 18 '12 at 19:20
I am extremely sorry for my writing and skipping that part with out mentioning in post, I must not in next posts. – Taxi Driver Dec 18 '12 at 19:23

## 1 Answer

You are right. To see that all group homomorphisms have the form you claim, is not completely trivla, as $\mathbb Q$ is not even finitely generated. But note that for $n\in\mathbb Z$ you have $f(n\cdot x)=n\cdot f(x)$, hence for $q=\frac ab$ with $b\ne 0$ it follows indeed that $$f(q)=\frac1b\cdot f(b\cdot q)=\frac1b\cdot f(a)=\frac1b\cdot a\cdot f(1)=q\cdot f(1).$$ From this your observations about a,b,c follow just as you write.

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Though I knew the proof but thank you very much for the response :) – Taxi Driver Dec 18 '12 at 19:15
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