15
$\begingroup$

I am looking for a group $G$ such that $\operatorname{Aut}(G)\cong \Bbb{Z}/8\Bbb{Z}$.

Obviously $\operatorname{Aut}(\Bbb{Z}_n)\ncong \Bbb{Z}/8\Bbb{Z}$ for any $n$. Also $\operatorname{Aut}(D_4)\cong D_4$, neither symmetric/alternating groups are of any help here. May be there is no group for which $\operatorname{Aut}(G)\cong \Bbb{Z}/8\Bbb{Z}$, 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 from 1957 by H. de Vries and A.B. de Miranda which says $C_8$ does not occur as Aut group of any torsion-free abelian group. So this settles the question.

$\endgroup$
1
  • 2
    $\begingroup$ $G/Z(G)$ injects into $Aut(G)$, so we can deduce that $G/Z(G)$ is cyclic. It is well known that if $G/Z(G)$ is cyclic, then $G$ is abelian. Therefore you can restrict to the case of abelian groups. Doesn't look likely at first sight. $\endgroup$ Commented Jun 30, 2015 at 20:37

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This only says that $G$ is abelian, and I know no finite abelian has Aut$(G)$ isomorphic to $\Bbb{Z}_8$, but we cannot say anything about infinite abelian groups here $\endgroup$ Commented Jun 30, 2015 at 20:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .