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A question I'm struggling with is to prove or disprove that the automorphism group of a finite cyclic group must be cyclic. I think the statement is false but have not been able to come up with a counterexample.

Thank you for any assistance you can provide.

qwerty
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    The automorphism group of $C_n$ is the multiplicative group of the units of the ring $\Bbb Z_n$. – Angina Seng Nov 18 '19 at 04:24
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    This is false, and the smallest counterexample occurs with the cyclic group of order 8, whose automorphism group is non-cyclic. Given this info, can you make any more progress on this problem? – cnpJj2dwc Nov 18 '19 at 04:24
  • groupprops.com claims "For a finite cyclic group of order n, the automorphism group is of order $\phi(n)$ where $\phi$ denotes the Euler totient function. Further, the automorphism group is cyclic iff n is 2,4, a power of an odd prime, or twice a power of an odd prime. In particular, for a prime p, the automorphism group of the cyclic group of order p is the cyclic group of order p - 1." I would compute the automorphism group for one of the non-examples as your counter example.https://groupprops.subwiki.org/wiki/Automorphism_group_of_a_group – Dominic Reiss Nov 18 '19 at 04:26
  • Try a cyclic group of order $8$. In general, one can show that $\text{Aut}(\mathbb{Z}_n)$ is isomorphic to the multiplicative group of $Z_n$ (just check for appropriate places to send generators). – user328442 Nov 18 '19 at 04:26

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For an automorphism, a generator has to go to a unit (a generator). Thus $\operatorname {Aut}(\Bbb Z_n)\cong \Bbb Z_n^×$.

Next, $\Bbb Z_n^×$ is cyclic iff $n$ is $1,2,4,p^k$ or $2p^k$, where $p$ is an odd prime. This is a little less obvious.

But you can check, for instance, that $\Bbb Z_8^×\cong \Bbb Z_2×\Bbb Z_2$ is not cyclic.