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Can anyone give me an example of:

An infinite abelian but non-cyclic group whose automorphism group is cyclic.

edited Sep 2 2011 at 9:16

asked Sep 1 2011 at 14:09

Chandrasekhar
2,060●1●11●28

In fact, there are non-cyclic abelian groups of every possible (infinite) cardinality satisfying the conditions. Of course, they cannot be finitely generated. – Steve D Sep 1 2011 at 15:58

Graham Leuschke · Accepted Answer · 2011-09-01 14:22:28Z UTC

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Let $G$ be the (non-finitely generated) additive group of rational numbers with square-free denominators. Then the only automorphism of $G$ is negation.

From here.

answered Sep 1 2011 at 14:22

Graham Leuschke
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@Graham: Thanks a lot. – Chandrasekhar Sep 1 2011 at 16:00

Adam Przezdziecki · Answer 2 · 2013-01-10 11:03:23Z UTC

One has a very general statement in this direction:

For any cotorsion-free ring $A$ there exist abelian groups $G$ of arbitrarily large cardinalities such that $$A\cong\mathop{\rm End}(G)$$

In particular if you need $\mathop{\rm Aut}(G)$ to be cyclic then you take $A$ such that the units of $A$ form a cylic group (e.g. $A=\mathbb{Z}$). Even more generally, you can replace abelian groups with $R$-modules, where $R$ is any commutative, cotorsion-free ring. Then you obtain $A$ - any cotorsion-free $R$-algebra.

You will find various constructions of such groups in Goebel and Trlifaj "Approximations and Endomorphism Algebras of Modules" (2006 and much expanded edition 2012).

*cotorsion-free above means no nontrivial homomorphisms of the underlying groups: $\mathbb{Z}^\wedge_p\to A$ for all primes $p$.*

Example of an infinite abelian but non-cyclic group whose automorphism group is cyclic.

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