Prove there isn't a group $G$ that satisfy ${\rm Aut}(G) \cong \mathbb{Z}$ [duplicate]

Question: Prove there isn't a group $G$ that satisfy ${\rm Aut}(G) \cong \mathbb{Z}$.

I saw this similar question Prove that there is no such group $G$ which would satisfy ${\rm Aut}(G)\cong\Bbb Z_n,$ where $n$ is an odd integer..

But I can't really understand why $n=1$ is trivial according to that commentary. Can someone give me a hint/help?

Solution 1:

$\operatorname{Aut}(G)$ being cyclic implies that $G$ is abelian. Now the automorphism $f : G \to G$ defined by $f(x) = x^{-1}$ has order $\leq 2$ in $\operatorname{Aut}(G)$. Since $\mathbb{Z}$ has no elements of order $2$, $f = \operatorname{id}_G$. Thus, every element of $G$ has order $1$ or $2$. This implies that $G$ is isomorphic to the additive group of an $\mathbb{F}_2$-vector space $V$, and it follows that $\mathbb{Z} \cong \operatorname{Aut}(G) \cong \operatorname{GL}(V)$. Since $\operatorname{GL}(V)$ is abelian, $\dim(V) \leq 1$. Then $\operatorname{GL}(V) \cong \operatorname{GL}_n(\mathbb{F}_2)$ for some $n \leq 1$, and thus $\operatorname{GL}(V)$ is finite. This contradicts the prior deduction that $\operatorname{GL}(V) \cong \mathbb{Z}$.