(Why) Is there no analogue to the classification of finitely generated abelian groups for abelian groups?

Question

Every finitely generated abelian group is a direct sum of cyclic groups. Does this hold for all abelian groups in general? If not, what fails?

Answer 1

No, it is badly false and the classification of abelian groups is much more complicated. For example $\mathbb{Q}$ is not a nontrivial direct sum in any way.

Exercise 1: Prove that if an abelian group $A$ is a nontrivial direct sum $B \oplus C$ then the endomorphism ring $\text{End}(A)$ contains a nontrivial idempotent (an element $e \neq 0, 1$ satisfying $e^2 = e$). Deduce that if $\text{End}(A)$ contains no nontrivial idempotents then $A$ is not a nontrivial direct sum.

Exercise 2: Prove that $\text{End}(\mathbb{Q}) \cong \mathbb{Q}$. Deduce that $\mathbb{Q}$ is not a nontrivial direct sum.

Exercise 3: For a prime $p$, the Prüfer $p$-group $\mu_{p^{\infty}}$ is the group of all $p$-power roots of unity, or equivalently the quotient $\mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$. Prove that $\text{End}(\mu_{p^{\infty}}) \cong \mathbb{Z}_p$ is the ring of $p$-adic integers. Deduce that $\mu_{p^{\infty}}$ is not a nontrivial direct sum.

In a positive direction you can see the Prüfer theorems, and you can also see Ulm's theorem.

In a negative direction, even the classification of countable torsion-free abelian groups is apparently hopeless in the following sense:

The classification problem for countable torsion free abelian groups is as complicated as that for arbitrary countable structures.