# Group homomorphisms $\mathbb{Q}\to\mathbb{R}$

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?

• What are $Q$ and $R$? – Daniel Aug 17 '15 at 16:55
• additive group of rationals and additive group of all reals, i think. – Foggy Aug 17 '15 at 16:57
• I suppose $Q$ and $R$ mean Rational and Real numbers. The problem is whether those are seen as groups or fields or something else. – Jose Paternina Aug 17 '15 at 16:59
• @JosePaternina OP said in the question he/she is looking for group homomorphisms. – Silvia Ghinassi Aug 17 '15 at 17:01
• @SilviaGhinassi you're right, sorry. – Jose Paternina Aug 17 '15 at 17:06

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 $f(\frac{1}{2}) = \frac{x}{2}$.
• 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