Question is to prove that :
A finite group whose only automorphism is identity map must have order at most $2$.
What i have tried is :
As any automorphism is trivial, so would be inner automorphism
i.e., each map for fixed $g\in G $ with $\eta : G\rightarrow G$ taking $h$ to $ghg^{-1}$ is trivial.
Thus, $ghg^{-1}=g$ i.e., $gh=hg$ for all $g\in G$ and $h\in G$ which would say that $G$ is abelian.
So, I would have that $G$ is finite abelian group.
Now, As $G$ is abelian, the map $g\rightarrow g^{-1}$ is an automorphism.
But only automorphism is identity map, so we would have :
$g=g^{-1}$ i.e., $g^2=e$ for all $g\in G$
So, I would have that $G$ is group with each element of order $2$.
Combining with previous result I would have :
$G$ is a finite abelian group in which each element is of order $2$
I am not able to conclude anything more than this….
A kind of cheating would give something very close :
As group is finite abelian which has each element with order $2$, It should be :
$\mathbb{Z}_2\times \mathbb{Z}_2$ Or
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or
$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or something very similar to this.
For first group $\mathbb{Z}_2\times \mathbb{Z}_2$ automorphism group is general linear group of order $2$ with entries from $\mathbb{Z}_2$ which is not trivial. So, this should not be the required group.
This would hold for similar cases
So, I feel that i am on right path but i need some help to make it more clear.
Thank you 🙂
Best Answer
A (finite) abelian group with $g^2=e$ for all $g\in G$ is in fact a (finite dimensional) $\Bbb Z_2$-vector space, and then its automorphisms correspond to invertible matrices over $\Bbb Z_2$, so...