September 25th, 2007, 09:53 AM  #1 
Newbie Joined: Sep 2007 Posts: 9 Thanks: 0  question on groups
i think i have spent to much time looking at this problem, and i just cant figure it out. i am probably just overlooking something obvious. So here it is. Let b be an element in some group G, and suppose that b has finite order m. Let c=b^k for some k. Prove that the order of c divides m. 
September 25th, 2007, 10:44 AM  #2 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7 
Well, you have that c^m=b^(km)=(b^m)^k=1 (the unit element). This means that the cyclic subgroup <c> generated by c has an order which is a divisor of m.


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