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Given a finite group $G$, how can I check whether a finite group $H$ exist, such that $\operatorname{Aut}(H)$ is isomorphic to $G$ ?

Here

http://groupprops.subwiki.org/wiki/Automorphism_group_of_a_group

some automorphismgroups are mentioned. We can see, that for $n\ne 2$ and $n\ne 6$, $S_n$ is equal to its own automorphismgroup.

For every $n$ for which $Z_n^*$ is cyclic , the cyclic group of order $\phi(n)$ is the automorphismgroup of $Z_n$.

But in general, the problem seems to be very difficult.

user1729
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Peter
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    Possible duplicate of http://math.stackexchange.com/questions/253936/is-every-group-the-automorphism-group-of-a-group. – lhf Apr 13 '16 at 19:00
  • For every $n;$ $U(n)$ is cyclic?$ – Upstart Apr 13 '16 at 19:01
  • @Ihf I think the question you mention deals with arbitary groups. – Peter Apr 13 '16 at 19:05
  • @Upstart No, $Z_n^$ is only cyclic in speacial cases. I did not claim that $Z_n^$ is always cyclic. I only mentioned the automorphismgroup under the assumption that it is. – Peter Apr 13 '16 at 19:07
  • You can distil my answer in lhf's link to prove that no cyclic group of odd order occurs as $\operatorname{Aut}(H)$ where $H$ is finite. (If $\operatorname{Aut}(H)$ is cyclic and non-trivial then $H$ is abelian, apply fundamental theorem for finitely generated abelian groups and notice that, as $\operatorname{Aut}(H)$ is non-trivial, there always exists an automorphism of order $2$.) – user1729 Apr 14 '16 at 08:09
  • (Incidentally, if instead of considering $\operatorname{Aut}(H)$ you consider $\operatorname{Out}(H):=\operatorname{Aut}(H)/\operatorname{Inn}(H)$ the problem becomes more tractable. For example, every finite group $G$ can be realised as the outer automorphism group $\operatorname{Out}(H)$ where $H$ is the fundamental group of a closed hyperbolic $3$-manifold.) – user1729 Apr 14 '16 at 08:15

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