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December 2nd, 2007, 07:57 PM  #1 
Member Joined: Sep 2007 Posts: 77 Thanks: 0  If a has order hk modulo n, then a^h has order k mod n. If a has order hk modulo n, then a^h has order k mod n. We are supposed to prove this and we can use Thm 8.3 to do it, which is: If the a has order k mod n, then a^h has order k/gcd(h,k) I know this is probably simple to most of you, but for some reason I am getting stuck when trying to write out the proof... Thanks 
December 2nd, 2007, 08:21 PM  #2 
Member Joined: Sep 2007 Posts: 77 Thanks: 0 
Nevermind, I figured it out. Sorry to waste anyone's time!


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