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January 15th, 2018, 02:11 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Order of an element in a Group
Q : An element 'a' in a group has order 100. Then what is the order of a^65 ? Property : if o(a) = n & p is prime to n then o(a^p) = n. Using the above property, removing common factors between 65 & 100, we get 13 & 20. Then o(a^65) would be 20 ? I just tried let me know how to solve above question if it's wrong. Thanks 
January 17th, 2018, 02:44 AM  #2 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography 
Good morning. Yes, your answer is correct. It is a result of the following property, which generalizes the one you mentioned in your post: Let $\displaystyle o(a)=n$ and let $\displaystyle m\in{\mathbb N}$. Then $\displaystyle o\left(a^m\right)=\dfrac nd$, where $\displaystyle d=\gcd(n,m)$. Last edited by skipjack; January 17th, 2018 at 04:52 PM. 
January 17th, 2018, 02:57 AM  #3 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 
Thank you so much for the response.
Last edited by skipjack; January 17th, 2018 at 04:53 PM. 
January 17th, 2018, 09:09 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 
Ould Yubba has given you the correct formula. The proof is straight forward: 

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