Suppose $G$ is a group. Does there always exist a group $H$, such that $\operatorname{Aut}(H)=G$, i. e. such that $G$ is the automorphism group of $H$?

EDIT: It has been pointed out that the answer to the above question is no. But I would be much more pleased if someone could give an example of such a group $G$ not arising as an automorphism group together with a comparatively easy proof of this fact.

  • $\begingroup$ @rondo9 Why does $\mbox{Aut}G$ simple imply that $G$ is abelian? What if $G$ has a trivial outer automorphism group? $\endgroup$
    – Alexander Gruber
    Commented Dec 8, 2012 at 20:02
  • $\begingroup$ Ah yes, my mistake. $\endgroup$
    – rondo9
    Commented Dec 8, 2012 at 20:46
  • $\begingroup$ This overflow link does not directly address what you are asking, but may be of some interes: mathoverflow.net/questions/37356/… $\endgroup$
    – Rankeya
    Commented Dec 8, 2012 at 22:29
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/1345052 $\endgroup$
    – Watson
    Commented Aug 17, 2016 at 10:00
  • $\begingroup$ By the way, "any finite group occurs as a certain quotient of Aut(G) for some finite p-group G" - arxiv.org/pdf/0711.2816.pdf $\endgroup$
    – Alex W
    Commented May 11, 2017 at 3:18

2 Answers 2


Theorem. The following cyclic groups cannot be the automorphism group of any group:

  1. The infinite cyclic group $C_{\infty}$ (also known as $\mathbb{Z}$), and
  2. Cyclic groups $C_{n}$ of odd order (also known as $\mathbb{Z}_n$ or $\mathbb{Z}/n\mathbb{Z}$).

The proof is relatively straight forward, with a subtlety at the end, and consists of two lemmata. I shall leave you to pin the lemmata together and get the result. (Hint. what are the inner automorphisms isomorphic to?)

Lemma 1: If $G/Z(G)$ is cyclic then $G$ is abelian.

Proof: This is a standard undergrad question, so I'll let you figure out the proof for yourself.

Lemma 2: If $G\not\cong C_2$ is abelian then $\operatorname{Aut}(G)$ has an element of order two.

(Here, $C_2$ is the cyclic group of order two. Note that this group has trivial automorphism group.)

Proof: The negation map $n: a\mapsto a^{-1}$ is non-trivial of order two unless $G$ comprises of elements of order two. If $G$ consists only of elements of order two then, applying the Axiom of Choice, $G$ is the direct sum of cyclic groups of order two, $$G\cong C_2\times C_2\times\ldots$$ See this question for why. Finally, because $G\not\cong C_2$ there are at least two copies of $C_2$, and so we can switch them (and "switching" has order two).

The subtlety I mentioned at the start is the use of Choice in the proof of Lemma 2. If we do not assume Choice that it is consistent that there exists a group $G$ of order greater than two such that $\operatorname{Aut(G)}$ is trivial. This was (first) proven by Asaf Karagila in an answer to this MSE question.


Evidently this is false even if $H$ is required to be finite.

I think the argument would be very difficult if $G$ was allowed to be infinite, especially if $G$ was not finitely generated.

EDIT: Here is a cool somewhat related result.

  • $\begingroup$ Thanks for the references but, unfortunately, they are behind a paywall. Do you know any articles which one can find on a free platform (i. e. on arxiv)? $\endgroup$
    – Dominik
    Commented Dec 8, 2012 at 20:26
  • $\begingroup$ @Dominik Sorry, no I don't. I was able to find them at my university library. If you're a student, they will likely be accessible from yours, too. $\endgroup$
    – Alexander Gruber
    Commented Dec 12, 2012 at 4:12
  • 5
    $\begingroup$ @Dominik: Check out the third paper on Inna Bumagin's webpage, here. She and Dani Wise proved that every countable group $Q$ is the outer automorphism group of a finitely generated group $N$. Their proof is, if I remember correctly, pretty self-contained. Their results about $N$ being residually finite when $Q$ is finitely presented can be ignored, as this is immediate from Wise's latest work (he has proven, among other amazing things, that every $C^{\prime}(1/6)$-group is residually finite, which solves the problem here.) $\endgroup$
    – user1729
    Commented Mar 14, 2013 at 13:28

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