It is easy to prove, e.g. here that any group $G$ with cyclic automoprhism group must be Abelian. Cyclic groups of order $\phi(p^n) = (p-1)p^{n-1}$ for $p \ne 2$ are obviously automorphism groups, as they are the automorphism groups of $C_{p^n}$. FTFAG proves these are the only possibilities for finitely generated Abelian groups. That link also proves that cyclic groups of odd order (other than $C_1$) cannot be automorphism groups. However, this obviously leaves the question open for many even orders, the smallest being $14$.

Other than these cases, I cannot prove anything more about possible orders of cyclic automorphism groups, or find any further theorems. This MathOverflow answer talks about non locally cyclic groups with cyclic automorphism groups, and suggests $C_2$, $C_4$ and $C_6$ may be the only possibilities. But all of these are of the form $C_{(p-1)p^m}$. I can't find anything talking about this question for general groups.

So I am wondering whether there are any groups with cyclic automorphism group not of order $(p-1)p^{m}$. So I would like to know any counterexample to this, or any proof that rules out any even order of cyclic groups.

  • $\begingroup$ Cyclic groups of order $2p^n$ for any odd prime $p$ have cyclic automorphism group. Among cyclic groups, those with cyclic automorphism group are precisely the groups of order $2$, $4$, $p^n$, and $2p^n$, for odd prime $p$. I believe these are the only abelian finite groups with cyclic automorphism group. $\endgroup$ Commented Mar 3, 2023 at 16:06
  • $\begingroup$ Of course, other abelian groups have cyclic automorphism group, e.g., the nonzero reals under multiplication.... $\endgroup$ Commented Mar 3, 2023 at 16:09
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    $\begingroup$ Could the people downvoting this question please explain what is wrong with it? $\endgroup$
    – Zoe Allen
    Commented Mar 3, 2023 at 18:31
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    $\begingroup$ @BenjaminDickman $6$ is of the form $(p-1)p^n$, with $p=3$, $n=2$. $\endgroup$ Commented Mar 3, 2023 at 18:41
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    $\begingroup$ @BenjaminDickman Sorry, I fixed it now. $\endgroup$
    – Zoe Allen
    Commented Mar 3, 2023 at 18:44

1 Answer 1


The situation about cyclic groups of automorphisms is as follows.

  1. If $G$ is an infinite periodic group, then its automorphism group is also infinite (R. Baer).

  2. If a cyclic group $A$ is the automorphism group of a torsion-free abelian group, then $A$ is of order $2$, $4$ or $6$ (J.T. Hallett and K. A. Hirsch)

  3. If a cyclic group $A$ is the automorphism group of a infinite abelian group, then $A$ is of order $2$, $4$ or $6$ (this follows from 1, 2).

  4. If a cyclic group $A$ is the automorphism group of a finite abelian group, then $A$ is of order $p^s(p-1)$ for some odd prime $p$ and integer non-negative $s$ (this follows from the fundamental theorem of finite abelian groups).

  5. By the way, I note that the list of those cyclic groups which can be automorphism groups of topological groups is much wider. For a complete list see here

Addition to item 3.

Let $G$ be an additive infinite abelian group with torsion part $T\neq G$. If $T\neq0$, then by Corollary 2.3 from L.Fuchs $G$ has a cocylic direct summand $G'$, that is $G=G'\oplus P$ where $P$ is isomorphic to $\mathbb{Z}(p^k)$ for some prime $p$ and for some $k\in\mathbb{N}\cup\{\infty\}$. If $|P|>2$, then $\operatorname{Aut}(G)$ contains a noncyclic group of order $4$. The rest is obvious.

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    $\begingroup$ Is it obvious that 3 follows from 1 and 2? $\endgroup$ Commented Mar 4, 2023 at 13:43
  • $\begingroup$ Which groups does statement $N_1$ apply to? I thought it only applied to torsion-free Abelian groups? in which case it wouldn't tell you anything that 2 doesn't. $\endgroup$
    – Zoe Allen
    Commented Mar 4, 2023 at 21:09
  • $\begingroup$ Also was $N_1$ meant to specifiy an exception for $p^s = 3$? It allows for order $6$ and $12$ cyclic groups doesn't it? $\endgroup$
    – Zoe Allen
    Commented Mar 4, 2023 at 21:09
  • $\begingroup$ @ZoeAllen I was somewhat confused by Jeremy Rickard's question and unsuccessfully started correcting my text. In fact, it's simpler than that. I supplemented my answer and removed the reference to $N_1$, which is unnecessary. $\endgroup$
    – kabenyuk
    Commented Mar 5, 2023 at 11:24

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