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

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 $$1$$) .

• One way to proceed is to search the site to see if the question has been asked before. Try here for example. Dec 31, 2020 at 9:28
• Does this answer your question? Cyclic Automorphism group Dec 31, 2020 at 9:29
• @ArsenBerk I have edited the question to include my effort...can you please look into it? Dec 31, 2020 at 9:34
• @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? Dec 31, 2020 at 9:36
• Please try and put some effort into solving this problem yourself. What happens when $|G|=4$ for example? Dec 31, 2020 at 11:49

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 $$\phi$$?
• 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. Dec 31, 2020 at 9:37
• @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. Dec 31, 2020 at 10:28
• @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). Dec 31, 2020 at 10:42