7

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.

Zoe Allen
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  • 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. – Arturo Magidin Mar 03 '23 at 16:06
  • Of course, other abelian groups have cyclic automorphism group, e.g., the nonzero reals under multiplication.... – Arturo Magidin Mar 03 '23 at 16:09
  • @ArturoMagidin the nonzero reals under multiplication are isomorphic to $C_2 \times \mathbb{R}$ which has a much larger automorphism group. And I am not asking which groups have cyclic automorphism group. I am asking which cyclic groups are automorphism groups. – Zoe Allen Mar 03 '23 at 16:29
  • "with cyclic The cyclic..." Can you fix that first sentence? Also, not all groups of order $\phi(p^n)$ are automorphisms of a cyclic group, only the cyclic group of order $\phi(p^n).$ – Thomas Andrews Mar 03 '23 at 16:33
  • True on reals, sorry. Yes, iI know your question. The point is that we know exactly what those automorphism groups are, so if you know all finite groups that have cyclic automorphism group, that tells you exactly which cyclic groups occur as automorphism groups (of finite groups). – Arturo Magidin Mar 03 '23 at 16:40
  • For nontrivial divisible groups, the automorphism group is never cyclic (you either have a direct summand isomorphic to $\mathbb{Q}$ or one to $\mathbb{Z}_{p^{\infty}}$, which gives an automorphism group that contains either the nonzero rationals, or the $p$-adic integers, neither of which is cyclic. Every abelian group can be written as a direct sum of a divisible group and a reduced group, so that you are left with "only" reduced infinite groups. – Arturo Magidin Mar 03 '23 at 17:30
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    Could the people downvoting this question please explain what is wrong with it? – Zoe Allen Mar 03 '23 at 18:31
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    @BenjaminDickman $6$ is of the form $(p-1)p^n$, with $p=3$, $n=2$. – Arturo Magidin Mar 03 '23 at 18:41
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    @BenjaminDickman Sorry, I fixed it now. – Zoe Allen Mar 03 '23 at 18:44
  • Could the people downvoting this question please explain what is wrong with it? Possibly not. (I guess we'll see.) Unfortunately, a large percentage of down-votes simply go unexplained. Some of them (PSQs, for example) are obvious and probably don't require explanation, but this clearly isn't one of them. FWIW I definitely don't see anything in this question worth a down-vote. – Brian Tung Mar 03 '23 at 23:27
  • FTFAG? ${}{}{}$ – Mariano Suárez-Álvarez Mar 04 '23 at 10:19
  • @MarianoSuárez-Álvarez Fundamental Theorem of Finite [or Finitely-generated] Abelian Groups – Zoe Allen Mar 04 '23 at 13:31

1 Answers1

3

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.

kabenyuk
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    Is it obvious that 3 follows from 1 and 2? – Jeremy Rickard Mar 04 '23 at 13:43
  • 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. – Zoe Allen Mar 04 '23 at 21:09
  • 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? – Zoe Allen Mar 04 '23 at 21:09
  • @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. – kabenyuk Mar 05 '23 at 11:24