Q:

Finding all groups with $\text{Aut}(G)=\{1\}$

Possible Duplicate:
$|G|>2$ implies $G$ has non trivial automorphism 

I am doing this exercise:

Find all groups $G$, with $\text{Aut}(G)=\{1\}$.

What has been clear to me is the group $G$ should be abelian group. Because we will have $G=Z(G)$ and I see that at least all $\phi_g(x)=g^{-1}xg$ are just identity. Any help is appreciated! 

A:

All the inner automorphisms of $G$ are trivial, so $G$ is abelian.
But for an abelian group $G$, the map $g\mapsto g^{-1}$ is an automorphism. So this must be the identity map, i.e. every nonidentity element has order $2$.
In other words, $G$ is a vector space over $\mathbb{Z}/2\mathbb{Z}$. Standard linear algebra techniques show that any space of dimension at least $2$ has a non-trivial automorphism. Thus $G$ must have dimension $0$ or $1$, i.e. $G$ is the trivial group or cyclic of order $2$.

