Problem 5.2.16 (No finitely generated abelian group is divisible)

Proof. Let $G$ be a finitely generated abelian group. Assume for the sake of contradiction that $G$ is divisible. By Theorem 3,

By definition of a divisible group, $G$ is not trivial and so $r>0$ or $s>0$.

By Exercise 2.4.20, every factor group is divisible. If $s>0$ then $Z_{n_1}$ is divisible, but this is impossible by Exercise 2.4.19: no finite abelian group is divisible.

Therefore $s=0$ and $G\simeq {\mathbb{Z}}^r$. Using Exercise 2.4.20 anew, $\mathbb{Z}$ would be divisible, but this is false, since $1$ is not the $5^{\mathrm{th}}$ multiple of some element of $\mathbb{Z}$.

This contradiction shows that no finitely generated abelian group is divisible. □