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 Dec 8 '12 at 20:02
  • $\begingroup$ Ah yes, my mistake. $\endgroup$ – rondo9 Dec 8 '12 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 Dec 8 '12 at 22:29
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    $\begingroup$ Related: math.stackexchange.com/questions/1345052 $\endgroup$ – Watson Aug 17 '16 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 May 11 '17 at 3:18

Theorem: The infinite cyclic group, written $C_{\infty}$, cannot be the automorphism group of any group.

Also known as $\mathbb{Z}$ (under addition), this is the canonical example of such a group (and it seems my proof works for all cyclic groups of odd order). The proof is easy, 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 $G$ 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). Note: This automorphism depends on the Axiom of Choice (this was pointed out by Arturo Magidin in an answer to a question on the group pub forum mailing list).


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 Dec 8 '12 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 Dec 12 '12 at 4:12
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    $\begingroup$ @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 Mar 14 '13 at 13:28

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