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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.

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<pedantry>The plural of 'criterion' is 'criteria'.</pedantry> –  HJRW Aug 12 '10 at 14:50
Typos now fixed. –  Jack Schmidt Aug 12 '10 at 15:23
Added more description to the title. –  Cam McLeman Aug 12 '10 at 16:31
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2 Answers

up vote 16 down vote accepted

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:

  • Szele, T.; Szendrei, J. "On abelian groups with commutative endomorphism ring." Acta Math. Acad. Sci. Hungar. 2, (1951). 309–324 MR51835 DOI:10.1007/BF02020735

This papers also gives more complicated examples of mixed groups with commutative endomorphism ring.

The mixed case was completed in:

  • Schultz, Ph. "On a paper of Szele and Szendrei on groups with commutative endomorphism rings." Acta Math. Acad. Sci. Hungar. 24 (1973), 59–63. MR316598 DOI:10.1007/BF01894610

This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").

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Thanks! I figured that it was rather unlikely that the general case would be simple, but I am surprised how easily the finitely abelian groups with commutative endomorphism rings are classified. –  Daniel Miller Aug 12 '10 at 20:18
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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.

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