Q. Show that there is no finite group $G$ such that $Aut(G)$ is isomorphic to $\mathbb{Z}_p$ for an odd prime $p$.
Attempt: If there is such a finite group $G$, then $G$ is abelian as $Aut(G)$ is cyclic. If there is an element in $G$ having order more than $2$, then the inversion automorphism has order $2$, which leads to a contradiction. Then we are only left with the option that $G$ is a finite abelian group having all elements with order at most $2$. Now how to proceed further without using advanced results of the Group Theory.
