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:
- 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?
Best Answer
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?