Assume that a is an element of order n in a group G. Prove that m and n are relatively
prime if and only if a^m has order n.
Thanks for your help...
You don't need to prove anything, this follows from the definitions: If $\displaystyle a$ is an element of order $\displaystyle n$ in a group and $\displaystyle m$ is a positive integer, then $\displaystyle <a^k>=<a^{gcd(n,m)}>$ and $\displaystyle |a^m|=\frac{n}{gcd(n,m)}$. So if n and m are relatively prime then $\displaystyle |a^m|= \frac{n}{gcd(m,n)}=\frac{n}{1}=n$.
To see why that definition holds, let d=gcd(n,m), clearly $\displaystyle (a^d)^{n/d}=a^n=e$, so that |$\displaystyle a^d| \leq n/d$. Also if $\displaystyle i$ is a positive integer less than n/d. then $\displaystyle (a^d)^i \ne e$ by definition of |a|. And since d=gcd(n,m) you get $\displaystyle |a^m|=|<a^m>|=|<a^{gcd(n,m)}>|=|a^{gcd(n,m)}| = \frac{n}{gcd(n,m)}$.