Let $G$ be a finite group. Under which conditions on $G$ is the automorphism group $\text{Aut }G$ cyclic? More precisely, does "$G$ is abelian" or "$G$ is cyclic" imply "$\text{Aut }G$ is cyclic"?

  • $\begingroup$ Well if you want a sufficient condition: $G=\{0\}$. A bit less trivial, $G=\mathbb{Z}/(p)$. $\endgroup$ – J.R. Jan 30 '14 at 13:57
  • $\begingroup$ @TooOldForMath That is not true for all $n$. $\endgroup$ – Tobias Kildetoft Jan 30 '14 at 13:59
  • $\begingroup$ @TobiasKildetoft Sorry, my bad. $\endgroup$ – J.R. Jan 30 '14 at 14:02

In fact, $\operatorname{Aut}(G)$ is cyclic if and only if $G$ is cyclic of order $n$ where $n$ is either $4$ or of the form $2^mp^k$ where $m$ is $0$ or $1$ and $p$ is an odd prime.

That $\operatorname{Aut}(G)$ is cyclic in these cases follows since it is will then be isomorphic to the multiplicative group of the integers mod $n$, which is cyclic in precisely the above mentioned cases.

If on the other hand $\operatorname{Aut}(G)$ is cyclic, then clearly $G$ is abelian, as $G/Z(G)$ is otherwise a non-cyclic subgroup.

But if $G = H\times K$ then $\operatorname{Aut}(H)\times \operatorname{Aut}(K)$ is naturally a subgroup of $\operatorname{Aut}(G)$ so if this is to be cyclic, then the orders of these must be coprime, and hence the order of $G$ is indeed as described above (or possible a larger power of $2$). The same argument also shows that the $p$-Sylow must be cyclic, so the only case left is that of some larger power of $2$. But now once again the same argument shows that $G$ would have to be a direct product of some number of copies of $C_2$, and we know that these groups do not even have abelian automorphism group.


This is only a sketch of a solution but the details can be worked out.

If $G$ is abelian then you can write $G=\prod_{i=1}^{n} \mathbb{Z}/(p_i^{r_i})$. It is easy to show that $\textrm{Aut(G)}=\prod_{i=1}^{n} \textrm{Aut}(\mathbb{Z}/(p_i^{r_i}))$ (Hint: Consider the order of the elements).

It is enough to consider the groups $\mathbb{Z}/(p^{r})$. This is computed here. For odd $p$, $\textrm{Aut}(\mathbb{Z}/(p^{r}))\cong \mathbb{Z}/(\varphi (p^{r}))$. The product result a cyclic group if and only if the orders are coprime. This is not an easy condition but for specific examples you can decide if you have a cyclic automorphism group.

There is also some kind of converse. See this link.


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