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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.

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@rondo9 Why does $\mbox{Aut}G$ simple imply that $G$ is abelian? What if $G$ has a trivial outer automorphism group? – Alexander Gruber Dec 8 '12 at 20:02
Ah yes, my mistake. – rondo9 Dec 8 '12 at 20:46
This overflow link does not directly address what you are asking, but may be of some interes:… – Rankeya Dec 8 '12 at 22:29
up vote 14 down vote accepted

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).

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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.

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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)? – Dominik Dec 8 '12 at 20:26
@Dominik Sorry, no I don't. If you're a student they might be accessible from your university library; I was able to find them at mine. – Alexander Gruber Dec 12 '12 at 4:12
@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.) – user1729 Mar 14 '13 at 13:28

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