Thread: order of elements in a group View Single Post
 Hurkyl, your genuis quite disgusts me. In any event, I was able to complete the proof. Thanks so much for your help! You weren't by any chance born knowing this, were you? In case you're interested, you have to instead compute (ab)^q=e^q. Once you get b^(qm)=e, you use the fact that b has order n and that n,m are relatively prime to show n|q. Similarly, you can show m|q. And so q is a common multiple of m,n, and so mn|q.