Proof that there's no group $G$ such that $\operatorname{Aut}(G)\cong Q_8$ I know that there is no $G$ such that $Q_8 \cong\operatorname{Aut}(G) $, but I can't find a detailed proof of this fact. 
Could someone provide me with a clear answer?
Thanks!
 A: The proper subgroups of $Q_8$ are all cyclic. So $G/Z(G)\cong \operatorname{Inn}(G)$ is either cyclic or $Q_8$ itself. If $G/Z(G)$ is cyclic then $G$ is abelian. 
In either case, $G$ is the direct product of a Sylow 2-subgroup $P$ and an abelian 2'-group $H$. If $|H|$ and $|K|$ are coprime then $ \operatorname{Aut}(H\times K)\cong\operatorname{Aut}(H) \times\operatorname{Aut}(K)$ and so $G$ must be a $p$-group for some prime $p$.
All nilpotent groups of order greater than $2$ have an outer automorphism and so $G$ must be abelian. 
If $G$ is a non-cyclic abelian group of order $p^n, n>2,$ then  $|G|$ divides $|\operatorname{Aut}(G) |$. This leaves just a few relatively simple cases to consider. 
Some other contributors may be able to provide you with a more self-contained proof than this one. The results quoted are from Some Finite Groups Which Are Rarely Automorphism Groups: II D. MacHale Proceedings of the Royal Irish Academy. Section A: Mathematical and Physical Science Vol. 83A, No. 2 (1983), pp. 189-196. 
