Prove that $|a^m|=n|$ giving gcd(m,n)=1 and $|a|=n

Question

We have a group with a in the group and we're given |a|=n and m being a positive integer s.t gcd(m,n)=1 then prove $|a^m|=n $??

Answer 1

Given $|a|=n$ and gcd$(m,n)=1$ we know that $a^m \neq e$ (e being the identity element). Note that $(a^m)^n$ = $a^{mn} = a^{nm} =(a^n)^m = e^m=e$. So we know that $|a^m| \ \big| \ n $. Suppose $|a^m|<n$. Then there should exist $k<n$ such that $(a^m)^k=e$. However, $(a^m)^k=e \implies (a^m)^k = a^{mk} = a^n \implies m|n \implies gcd(m,n)=m$, so we must have that $|a^m| = n$