$p$ be a prime number , $G$ be a group of order $p+1$ such that $\exists f \in Aut(G)$ of order $p$ , then is $G$ abelian and has prime exponent?

Question

Let $p$ be a prime number , $G$ be a group of order $p+1$ such that there exists $f \in Aut(G)$ of order $p$ ; then is it true that $G$ is abelian and there exist a prime number $q$ such that $g^q=e , \forall g \in G$ ?

Answer 1

Yes. Note that $f$ permutes the $p$ nontrivial elements of $G$, so it must form a $p$-cycle, i.e., it cycles all nontrivial elements of $G$. Let $q$ be a prime factor of $p+1$ and let $g \in G$ be an element of order $q$, which exists by Cauchy's theorem. Then $\{f^i(g) \mid 0 \le i < p\} = G-\{e\}$ implies that every nontrivial element of $G$ has order $q$, i.e, the exponent of $G$ is $q$.

By Cauchy's theorem, this implies $|G| = q^k$ for some natural number $k$, so $Z(G) \neq \{e\}$. Let $h \in Z(G)$ be a nontrivial element. Then, once again, $f$ carries $h$ to all other nontrivial elements of $G$, so we get $Z(G) = G$.