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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?

Shaun
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geep
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    In the question you cite, they're trying to show that there is no $G$ such that $Aut(G) \cong \mathbb{Z} / n \mathbb{Z}$, which is very different than what you're trying to show. This theorem is actually not true for $n = 1$, since the zero group's group of automorphisms is $0$. Note that $\mathbb{Z} / 1 \mathbb{Z} \cong 0$. – Mark Saving Dec 02 '21 at 00:46
  • This question is fully answered here – kabenyuk Dec 02 '21 at 03:34
  • @geep I've closed this as a duplicate, but do you understand MarkSavings explanation of the $n=1$ case? (The duplicate doesn't answer the $n=1$ question, but you are going down the wrong path with $n=1$ - really you want $n=\infty$, i.e. $\mathbb{Z}=\mathbb{Z}{\infty}$, for whatever "$\mathbb{Z}{\infty}$" means.) – user1729 Dec 02 '21 at 07:51

1 Answers1

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$\operatorname{Aut}(G)$ 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 $\operatorname{GL}(V) \cong \mathbb{Z}$.

diracdeltafunk
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