If $p$ is an odd prime then prove that there cannot exist a finite group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_p$.

Can anyone tell me how to proceed in this question?

Here is my attempt:

If ${\rm Aut}(G)$ is cyclic then ${\rm Inn}(G)$ is also cyclic which implies $G/Z(G)$ is cyclic which implies $G$ is abelian. If $G$ is abelian then for $\phi \in{\rm Aut}(G)$ such that $\phi(g)=g^{-1}$. then $\phi(\phi(g)) = \phi(g^{-1}) = g$. Hence $\phi$ is of order $2.$

Hence the order of ${\rm Aut}(G)$ must be a multiple of 2 and hence cannot have prime order.

Edit:- I cannot seem to proceed when $\phi$ becomes just the identity mapping. In that case every element of the group has order $2$ (except for the identity element of order $1$) .

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    $\begingroup$ One way to proceed is to search the site to see if the question has been asked before. Try here for example. $\endgroup$
    – Derek Holt
    Dec 31, 2020 at 9:28
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    $\begingroup$ Does this answer your question? Cyclic Automorphism group $\endgroup$
    – ArsenBerk
    Dec 31, 2020 at 9:29
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    $\begingroup$ @ArsenBerk I have edited the question to include my effort...can you please look into it? $\endgroup$ Dec 31, 2020 at 9:34
  • $\begingroup$ @DerekHolt Yes thank you very much.....This was the final detail I was missing. Thanks for quick reply. I think it is because then $\phi$ fails to be a homomorphism...am I correct? $\endgroup$ Dec 31, 2020 at 9:36
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    $\begingroup$ Please try and put some effort into solving this problem yourself. What happens when $|G|=4$ for example? $\endgroup$
    – Derek Holt
    Dec 31, 2020 at 11:49

1 Answer 1


Hint: yes $G$ is abelian and consider the map $\phi$: $G \rightarrow G$, defined by $\phi(g)=g^{-1}$. Prove that this is an isomorphism. What is the order of $\phi$?

  • $\begingroup$ This is the correct argument but you have to say something more to cover the case when $g\mapsto g^{-1}$ is actually the identity. $\endgroup$ Dec 31, 2020 at 9:37
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    $\begingroup$ Yes I know, I leave that of course to @Arghyadeep etc. Remember, this is a Hint. $\endgroup$ Dec 31, 2020 at 9:39
  • $\begingroup$ @NickyHekster How do I proceed when $\phi$ is the identity mapping ? In that case all elemets of the group is of order $2$ except the identity element. $\endgroup$ Dec 31, 2020 at 10:28
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    $\begingroup$ @Arghyadeep Chatterjee: note that in this case $G=\mathbb{F}_2^r$ for some $r \geq 1$ and you can compute the automorphism group (eg show it’s not abelian). $\endgroup$
    – Aphelli
    Dec 31, 2020 at 10:42
  • $\begingroup$ @Mindlack I do not understand what F^r2 means. I am a noob in group theory. $\endgroup$ Dec 31, 2020 at 11:33

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