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September 25th, 2007, 09:53 AM   #1
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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.
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September 25th, 2007, 10:44 AM   #2
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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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