Aut$(G)\cong \Bbb{Z}_8$

I am looking for a group such that Aut$(G)\cong \Bbb{Z}_8$.

Obviously Aut$(\Bbb{Z}_n)\ncong \Bbb{Z}_8$ for any $n$. Also Aut$(D_4)\cong D_4$, neither symmetric/alternating groups are of any help here. May be there is no group for which Aut$(G)\cong \Bbb{Z}_8$, this also raises a question can any finite/finitely generated (why not infinite) group can be generated as an automorphism group of some $G$


Update- It is clear now that $G$ has to be abelian, and no finite abelian satisfy it. Also I found a paper by de Miranda which says $C_8$ does not occur as Aut group of any torsion free abelian group. So this settles the question.

Solutions Collecting From Web of "Aut$(G)\cong \Bbb{Z}_8$"

If such a group $G$ exists, then it is abelian and not finitely generated, and its only possible decomposition as a direct sum is $G\cong H\times\Bbb{Z}/2\Bbb{Z}$, if it has any such decomposition at all.


Let $G$ be a group such that $\operatorname{Aut}(G)\cong\Bbb{Z}/8\Bbb{Z}$. Then the conjugation action
$$\psi:\ G\ \longrightarrow\ \operatorname{Aut}(G):\ g\ \longmapsto\ (x\ \longmapsto gxg^{-1}),$$
has $\ker\psi=Z(G)$, the center of $G$. Its image is a subgroup of $\operatorname{Aut}(G)\cong\Bbb{Z}/8\Bbb{Z}$ and hence cyclic, so $G/Z(G)$ is cyclic. It follows that $G=Z(G)$, so $G$ abelian.

Suppose there are groups $H_1$ and $H_2$ such that $G\cong H_1\times H_2$. Then we have an injection
$$\operatorname{Aut}(H_1)\times\operatorname{Aut}(H_2)\ \longrightarrow\ \operatorname{Aut}(H_1\times H_2):\ (\varphi_1,\varphi_2)\ \longmapsto\ ((h_1,h_2)\ \longmapsto\ (\varphi_1(h_1),\varphi_2(h_2))),$$
where $\operatorname{Aut}(H_1\times H_2)\cong\Bbb{Z}/8\Bbb{Z}$, so without loss of generality $\operatorname{Aut}(H_2)=0$ and hence either $H_2=0$ or $H_2=\Bbb{Z}/2\Bbb{Z}$. Moreover this shows that $H_1$ is not a nontrivial direct sum of two groups.

If $H_1$ is finitely generated, then by the structure theorem for finitely generated abelian groups it is a finite direct sum of cyclic groups of infinite or prime power order. By the above $H_1$ is cyclic, and as $\operatorname{Aut}(\Bbb{Z})=\Bbb{Z}/2\Bbb{Z}$ we see that $H_1$ is finite cyclic of prime power order, say $p^k$. Then
$$8=|\operatorname{Aut}(H_1)|=\varphi(p^k)=p^{k-1}(p-1),$$
which shows that $p^k=2^4$, but $\operatorname{Aut}(\Bbb{Z}/2^4\Bbb{Z})\not\cong\Bbb{Z}/8\Bbb{Z}$. So $H_1$ is not finitely generated.