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In introduction to algebra we got the exercise:

Let $G$ be a group. Show that when $\operatorname{Aut}(G)$ is cyclic $G$ is abelian.

This doesn't make that much trouble. Denote the center (all commuting elements) with $Z$. Then $G/Z$ is isomorphic to $\operatorname{Int}(G)$ where $\operatorname{Int}(G)$ denotes the subgroup of inner automorphisms. As every subgroup of a cyclic group is cyclic we have $G/Z$ is cyclic and hence $G$ is abelian (As a group which is cyclic of the center is abelian).

So the question is:

Is there a non cyclic group with a cyclic automorphism group?

We discussed the question already in chat and I found (thanks to google) a solution there, but I would enjoy other Examples.

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  • $\begingroup$ I would like a torsion-free example $G$ that does not embed in $\mathbb{Q}$. $\endgroup$ – Jack Schmidt May 22 '13 at 21:26
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Whenever $G$ is finite and its automorphismus is cyclic we can already conclude that $G$ is cyclic.

Because as we already saw $G$ is abelian and finite, we can use the fundamental theorem of finitely generated abelian groups and say that wlog $G=\mathbb{Z}/p^k\mathbb{Z} \times \mathbb{Z}/p^j \mathbb{Z}$. But the automorphismgroup isn't abelian and hence isn't cyclic.

For non finite groups the implication isn't true.

The following is from this link and only slightly reworded.

Let $G$ be the subgroup of the additive group of rational numbers comprising those rational numbers that, when written in reduced form, have denominators that are square-free numbers, i.e., there is no prime number $p$ for which $p^2$ divides the denominator.

Then: The only non-identity automorphism of is the negation map, so the automorphism group is $\mathbb{Z}/2\mathbb{Z}$, and is hence cyclic.

The group $G$ is not a cyclic group. In fact, it is not even a finitely generated group because any finite subset of can only cover finitely many primes in their denominators. It is, however, a locally cyclic group: any finitely generated subgroup is cyclic.

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    $\begingroup$ There are uncountably many non-isomorphic but really-very-similar examples: for instance the "cube free numbers" consisting of all rational numbers whose denominator is not divisible by the cube of any prime. Again the endomorphism ring is the largest subring contained in it ($\mathbb{Z}$), and the automorphism group is the group of units ($\pm1$). $\endgroup$ – Jack Schmidt May 22 '13 at 20:14
  • $\begingroup$ ...and if $G$ is the Klein $4$-group?... $\endgroup$ – user1729 May 22 '13 at 20:20
  • $\begingroup$ @user1729: automorphism group is GL(2,2) = S3, nonabelian $\endgroup$ – Jack Schmidt May 22 '13 at 20:22
  • $\begingroup$ (@Dominic: It wouldn't hurt to put in a little more of the proof sketch in the finite case, as the automorphism group of a subgroup and/or quotient group can be much larger than the automorphism group of the original.) $\endgroup$ – Jack Schmidt May 22 '13 at 20:23
  • $\begingroup$ @JackSchmidt Oh. Whoopse! $\endgroup$ – user1729 May 22 '13 at 20:58
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$\mathbb{Q}^*$ is non cyclic, but the only automorphism is the identity.

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    $\begingroup$ I think switching $2$ and $3$ in the prime factorization of $a/b$ is an automorphism. $\endgroup$ – Jack Schmidt May 22 '13 at 20:29
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    $\begingroup$ ${\bf Q}^\times\cong {\bf Z}/2{\bf Z}\oplus{\bf Z}\oplus{\bf Z}\oplus{\bf Z}\oplus\cdots$ has uncountably large nonabelian automorphism group (in particular it includes the infinite symmetric group). $\endgroup$ – anon May 22 '13 at 20:33
  • $\begingroup$ Even though this automorphism alone isn't enough but inversion is another automorphism ($x\mapsto x^{-1}$). $\endgroup$ – Dominic Michaelis May 23 '13 at 7:27
  • $\begingroup$ I think, the only groups with identity automorphism group are $1$ and $\mathbb{Z}/2$. $\endgroup$ – RDK May 31 '13 at 8:40

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