$x \rightarrow x^n$ is a group automorphism of a finite abelian group G

Question

How do we prove that the map $\phi:G \rightarrow G$ defined by $\phi(x) = x^n$ for some $n \geq 0$ is a group automorphism of $G$ if $\gcd(|G|,n)=1$?

Answer 1

First, show that $\phi$ is a homomorphism. From there, proceed by contradiction:

If $\phi$ were not an automorphism, then $\phi$ could not be injective. That is, $x^n = y^n$ for some $x \neq y$. With some re-arranging, we'd have $(xy^{-1})^n = e$.

There is a contradiction here. Think Lagrange.