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December 2nd, 2007, 07:57 PM   #1
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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...

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December 2nd, 2007, 08:21 PM   #2
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Nevermind, I figured it out. Sorry to waste anyone's time!
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