Homomorphisms from $\mathbb Q$ to a finitely generated abelian group

Question

Let $G$ be a finitely generated abelian group. Prove that there is no non-trivial homomophism from $\mathbb Q$ to $G$.

I believe this has to do with divisibility. $\mathbb Q$ is divisible. The image of it under a group homomorphism must be divisible. If $G$ were finite, this would imply that the image is the trivial subgroup since any divisible finite group is trivial. But $G$ is not necessarily finite. What should the argument be like instead?

Answer 1

A finite generated abelian group $G$ is isomorphic to $\oplus_{i=1}^{i=n}\mathbb{Z}/n_i\mathbb{Z}$. Let $x\in G$, and $f:\mathbb{Q} \rightarrow G$ such that $f(1)=x, f(n.{1\over n}) =nf({1\over n}) =x$. Write $x=(x_1,...,x_n), x_i\in\mathbb{Z}/n_i\mathbb{Z} $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.

Answer 2

Since $G$ is f.g., it's sufficient to prove the result for $G = \mathbb{Z}$ and $G = \mathbb{Z}_{p^n}$. But neither of these groups contains a nontrivial divisible subgroup.

Answer 3

A nonzero divisible group is not finitely generated.

Consider a homomorphism $f\colon\mathbb{Q}\to G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $\pi\colon G\to G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $\pi\circ f\colon\mathbb{Q}\to G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $\pi\circ f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $\mathbb{Q}^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $\pi\circ f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.

No nonzero divisible group is finite.