TITLE: If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .
QUESTION [7 upvotes]: If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .
With the assumption, I dont know how to start the proof.
If there is no non-trivial automorphism, then there is only trivial automorpism, the identity morphism. But how can I show  $ g^2 = e $ for all $ g \in G $ with it?

REPLY [4 votes]: After the observations above, note that in case $G$ is finite (for infinite groups a similar reasoning applies), it must be an elementary abelian 2-group of rank say $n$. Then $Aut(G) \cong GL(n,2)$. Hence if $|Aut(G)| = 1$ then this can only be the case if $G$ is trivial, or $n = 1$ that is $G \cong C_2$, the cyclic group of order 2.