Let $G$ be a finite Abelian group and let $n$ be a positive integer that is relatively prime to $|G|$. Show that the mapping $a \rightarrow a^n$ is an automorphism of G.
I've proven everything except for one-to-one-ness. I start with $a^n = b^n$ but I've made no progress with showing that $a=b$.
Anyone have any ideas?
Best Answer
Since $n$ is coprime to $|G|$, there exists an integer $m$ such that $nm\equiv 1$ (mod $|G|$). Hence if $a^n=b^n$, then $$ a=a^{nm}=(a^n)^m=(b^n)^m=b^{nm}=b $$ so the map is injective.