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Find all the group homomorphisms from $(\mathbb{Q}, +)$ into $(\mathbb{R}, +)$.

My attempt:

If $\mathbb{Q}$ were a cyclic group, I could tell that any homomorphism will be determined by the image of generator. But here $\mathbb{Q}$ is not a cyclic group, so there's no generator. All one can say is that:

  1. if $f$ is a homomorphism then $f(0)=0$.

But this doesn't help me to solve this problem. So how should it be tackled?

Patrick Stevens
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Foggy
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1 Answers1

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Let $f: (\mathbb{Q}, +) \to (\mathbb{R}, +)$ be a group homomorphism. As you say, $f(0) = 0$.

What happens to $1$? Let's say $f(1) = x$. Now, this fixes all the naturals: $f(1+1) = f(1) + f(1) = 2x$, and so on, so $f(n) = nx$.

What happens to $\frac{1}{2}$, which is what comes to mind as the simplest non-integer rational? $x = f(\frac{1}{2} + \frac{1}{2}) = 2 f(\frac{1}{2})$, so $f(\frac{1}{2}) = \frac{x}{2}$.

Can you generalise that yourself?

Patrick Stevens
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  • let m/n be any rational, i think then f(n(m/n))=nf(m/n), this means f(m/n)=(mx)/n..? is that all so we determined one morphism by defining both naturals and rationals.. so, is there only one morphisms between these two groups? – Foggy Aug 17 '15 at 17:09
  • You're right in saying that $f(n \times \frac{m}{n}) = n f(\frac{m}{n})$, and so $f(\frac{m}{n}) = \frac{m x}{n}$. That means there is only one morphism after having fixed the image of $1 \in \mathbb{Q}$. How many morphisms does that mean there are in total between the two groups? – Patrick Stevens Aug 17 '15 at 17:25
  • @PatrickStevens do you mean that there are c number of morphisms between the groups? –  Feb 06 '18 at 04:34
  • @ThatIs Correct. – Patrick Stevens Feb 06 '18 at 08:38