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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 $1$) .

Mr.Gandalf Sauron
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1 Answers1

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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 $\phi$?

Nicky Hekster
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  • 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. – Andrea Mori Dec 31 '20 at 09:37
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    Yes I know, I leave that of course to @Arghyadeep etc. Remember, this is a Hint. – Nicky Hekster Dec 31 '20 at 09:39
  • @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. – Mr.Gandalf Sauron Dec 31 '20 at 10:28
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    @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). – Aphelli Dec 31 '20 at 10:42
  • @Mindlack I do not understand what F^r2 means. I am a noob in group theory. – Mr.Gandalf Sauron Dec 31 '20 at 11:33
  • Perhaps this helps: https://math.stackexchange.com/questions/17054/group-where-every-element-is-order-2 – Nicky Hekster Dec 31 '20 at 11:50
  • @Arghyadeep Chatterjee: $\mathbb{F}_2$ is $\mathbb{Z}/2\mathbb{Z}$. Let $e_i$ be the vector with zeroes everywhere except a $1$ at the $i$-th place. Show that the group has an automorphism of order $2$, which exchanges $e_1$ and $e_2$. – Aphelli Dec 31 '20 at 12:07