Let $G$ be a group. Show that if ${\rm Aut}(G)$ (the group of automorphism of $G$ ) is cyclic, then $G$ is abelian, and if $G$ is additionally finite, show that $G$ is cyclic.
I need the second part of the question.
For the first part, I use this lemma
Lemma Every subgroup of a cyclic group is cyclic.
Now for a group $G$, if ${\rm Aut}(G)$ is cyclic then $G$ is abelian.
Solution:
Let suppose $G$ is a group with $\operatorname{Aut}(G)$ cyclic. Then, considering the conjugation morphism $\gamma: G \rightarrow \operatorname{Aut}(G)$ sending $g$ to the inner automorphism $x \mapsto g x g^{-1}$, we have that $\operatorname{ker}(\gamma)=Z(G)$, so that $G / Z(G) \cong \operatorname{Im}(\gamma) \leq \operatorname{Aut}(G)$, and by previous point $G_{1}:=G / Z(G)$ is cyclic. Take a generator $t . Z(G)$ for $G_{1}$, by fixing a suitable $t \in G$. Then every element $x \in G$ can be written as $x=t^{n} c$, with $c \in Z(G)$ and $n \in \mathbb{Z}$. We now prove that $t \in Z(G)$, so that $Z(G)=G$ and $G$ is abelian. Let $x \in G$ and take the commutator $[x, t]=x t x^{-1} t^{-1}$. Writing down $x=t^{n} c$, we get $$ [x, t]=t^{n} c t c^{-1} t^{-n} t^{-1}=t^{n+1} c c^{-1} t^{-n-1}=1_{G} $$ because $c \in Z(G)$. Hence $t \in Z(G)=G$ and we are done.
Let's use this lemma.
Lemma. Every cyclic group is abelian.
Since ${\rm Aut}(G)$ is cyclic, we have that ${\rm Aut}(G)$ is abelian.
By first part, we have that $G$ is abelian.
Since $G$ is finite abelian and ${\rm Aut}(G)$ is abelian, I need to deduce that $G$ is cyclic but I am missing the argument.
I am thinking of If $G$ is finite and not cyclic, it has a factor $C_{p^r}$ $C_{p^s}$.
Could someone help me finish the argument?