Timeline for A group whose automorphism group is cyclic

6 events

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Sep 11, 2016 at 9:16	comment	added	Jeremy Rickard		@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 8, 2016 at 12:16	comment	added	H.Durham		@JeremyRickard Well of course $C_2\times C_2$ is a counterexample, unless you are assuming what j.p is in their comment.
Feb 24, 2015 at 14:37	comment	added	Jeremy Rickard		@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 14:21	comment	added	joro		Is there a finite solution?
Feb 24, 2015 at 12:56	vote	accept	W4cc0
Feb 24, 2015 at 12:44	history	answered	Jeremy Rickard	CC BY-SA 3.0