# A group whose automorphism group is cyclic

Is there an Abelian group $A$ which is not locally cyclic whose automorphism group is cyclic ?

This question was first posted here.

• As we may assume $2\cdot A\ne 0$ (all elements having order $2$ is easy), we know that taking the inverse is the unique automorphism of order $2$. In particular, the cyclic automorphism group is finite.
– j.p.
Feb 24, 2015 at 10:44

There's a construction of a rank two (and therefore not locally cyclic) abelian group with endomorphism ring $\mathbb{Z}$, and therefore automorphism group cyclic of order 2, in "On the cancellation of modules in direct sums over Dedekind domains" by L. Fuchs and F. Loonstra, Indagationes Mathematicae, Volume 74, (1971), 163-169 (link)
• @joro You mean a non-cyclic finite abelian group with cyclic automorphism group? No, any non-cyclic finite abelian group is of the form $A\times B$ with $A$ and $B$ non-trivial, and so its automorphism group contains the non-cyclic group generated by inverting elements of $A$ and/or $B$. Feb 24, 2015 at 14:37
• @JeremyRickard Well of course $C_2\times C_2$ is a counterexample, unless you are assuming what j.p is in their comment. Sep 8, 2016 at 12:16
• @H.Durham Yes, you're quite right. The argument in my comment doesn't apply to all finite abelian groups, and another argument is needed for those of the form $G=C_2^n\times C_{p^k}$. If $n>1$ (or $p^k=2$) then $\operatorname{Aut}(G)$ contains $\operatorname{Aut}(C_2\times C_2)$ which is non-cyclic. If $n=1$ then $p=2$ (or else $G$ is cyclic), so $G\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^k\mathbb{Z}$, and the automorphism fixing $(1,0)$ and mapping $(0,1)$ to $(1,1)$ doesn't commute with the automorphism sending $(1,0)$ to $(1,2^{k-1})$ and fixing $(0,1)$. Sep 11, 2016 at 9:16