Suppose G is a group. Does there always exist a group H, such that 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 isno. 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.

**Answer**

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

- The infinite cyclic group C∞ (also known as Z), and
- Cyclic groups Cn of odd order (also known as Zn or Z/nZ).

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≇ 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.

**Attribution***Source : Link , Question Author : Dominik , Answer Author : user1729*