Given an abelian group $G$, one can form the endomorphism ring $\mbox{End}(G)$ by letting $\alpha+\beta=\alpha(x)+\beta(x)$, and $\alpha\beta=\alpha(\beta(x))$, where $\alpha$ and $\beta$ are endomorphisms. Clearly, composition distributes over addition, and addition is commutative, so $\mbox{End}(G)$ is a ring. My question is: when is $\mbox{End}(G)$ commutative? Are there a nice set of criteria, or, if there is no such nice set of criteria, is there a nice class of abelian groups with commutative endomorphism rings.

Amongst the finitely generated abelian groups, those with commutative endomorphism ring are exactly the cyclic groups. Torsion abelian groups with commutative endomorphism rings are exactly the locally cyclic groups, that is, the subgroups of Q/Z. They were classified in:
This papers also gives more complicated examples of mixed groups with commutative endomorphism ring. The mixed case was completed in:
This paper indicates the difficulty of any classification of torsionfree abelian groups with commutative endomorphism rings, as Corner has shown that very large torsionfree abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones"). 


In the finitely generated case it follows by the structure theorem that the unique possibilities are $G=\mathbb{Z}$ and $G=\mathbb{Z}/({p_1}^{e_1})\oplus \mathbb{Z}/({p_2}^{e_2})\oplus\cdots\oplus \mathbb{Z}/({p_k}^{e_k})$, where $p_1,\dots,p_k$ are distinct primes. Edit: in a word, the unique possibility is that $G$ is cyclic, as remarked by Jack Schmidt below. 

