Group with Prime Order

Question

If $G$ is a group with prime order then does $|\text{Aut}(G)|=|G|$? I am unsure of how to start this and where to work from. How can I approach prove this? Or what are some counter examples?

Answer 1

Given that $G$ is cyclic whenever $|G|$ is prime, then an automorphism of $G$ is any homomorphism $\varphi:G\to G$ such that $\varphi(g)=h,$ where $g,h\in G$ are such that $\langle g\rangle=\langle h\rangle=G.$ Furthermore, one can prove that, so long as neither $g$ nor $h$ is the identity of $G,$ then such a homomorphism (which will be uniquely determined by $g$ and $h,$ given the fact that $G$ is cyclic) is readily an automorphism. Fixing any non-identity $g\in G,$ there are thus $|G|-1$ choices for $h$ that yield an automorphism, so that there are $|G|-1$ automorphisms of $G.$

Answer 2

No. A simple counterexample is $G=C_2$, which has a trivial automorphism group.

In general, the automorphism group of the cyclic group of order $n$ is the cyclic group of order $\phi(n)$ because an automorphism has to preserve generators and there are $\phi(n)$ generators.

Therefore, $ |Aut(G)|=|G|$ for a cyclic group $G$ is true only for the trivial group.