Recent content by Worded.Mouse

Could someone give me a hint?

Prove that ordda | ordma, when d|m. Some conditions are 1 ≤ d, 1 ≤ m, and gcd(a,d)=1. What I have so far: let x=ordma, which gives us ax\equiv 1 (mod m) \Rightarrow ax=mk+1 for some k\in[SIZE="3"]Z Let m=m'd. Then ax=mk+1=d(m'k)+1