I have to find all the $$ Hom_g(\mathbb{Q}, \mathbb{Z}) . $$
In the exercise, it doesn't say which are the operations for the groups, so I assumed, it had to be in both cases the regular sum. What I thought, and I'm not completely sure it's right, is that there only is one $$f\in Hom_g(\mathbb{Q}, \mathbb{Z}),$$ which is the trivial case, f(q)=0. I did the following:
Since $$f\in Hom_g(\mathbb{Q}, \mathbb{Z}).$$ it follows that, f(0)=0 and f(p+q)=f(p)+f(q). Suppose that exists, and $$p\in\mathbb{Q}$$ such that $$0<f(p)=n\in\mathbb{Z}$$ So, if n is odd, then n=f(p)=f(p/2 + p/2) = f(p/2) + f(p/2) , then f(p/2)=n/2 absurd. If n is not odd, then, let's say in it's prime composition, the power of 2 y m, so $$f(p/2^m)\notin\mathbb{Z}$$
Then, it has to be f(p)=0, for all p in Q. Am I doing something wrong?
Thanks in advance