# Is every group the automorphism group of a group?

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.

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

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

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 $$G/Z(G)G/Z(G)$$ is cyclic then $$GG$$ is abelian.

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

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

(Here, $$C_2C_2$$ is the cyclic group of order two. Note that this group has trivial automorphism group.)

Proof: The negation map $$n: a\mapsto a^{-1}n: a\mapsto a^{-1}$$ is non-trivial of order two unless $$GG$$ comprises of elements of order two. If $$GG$$ consists only of elements of order two then, applying the Axiom of Choice, $$GG$$ is the direct sum of cyclic groups of order two, $$G\cong C_2\times C_2\times\ldotsG\cong C_2\times C_2\times\ldots$$ See this question for why. Finally, because $$G\not\cong C_2G\not\cong C_2$$ there are at least two copies of $$C_2C_2$$, 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 $$GG$$ of order greater than two such that $$\operatorname{Aut(G)}\operatorname{Aut(G)}$$ is trivial. This was (first) proven by Asaf Karagila in an answer to this MSE question.