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December 3rd, 2012, 03:56 PM  #1 
Newbie Joined: Nov 2012 Posts: 6 Thanks: 0  Order of a Subgroup M
Hello, I'm working through a proof right now about a direct product group, and the proof uses something that the book hasn't proved before, and I'm having a hard time proving it myself. G is a finite abelian group of order mn, m and n are relatively prime if the set M = { x element of G  x^m=e} it states that M=m it seems there might not always be an element x : o(x) = m, So M might contain elements of orders dividing m, so M might not equal m. Any help would be appreciated, thank you! 
December 24th, 2012, 01:48 AM  #2 
Newbie Joined: Dec 2012 Posts: 7 Thanks: 0  Re: Order of a Subgroup M
This is not a one liner. It uses Sylow's theorem and the fact that any finite abelian group is the direct product of it's Sylow's subgroups. Each Sylow's pgroup is the direct product of cyclic subgoups of order . If you put all that together, G is just the direct product of a group of order m and a group of order n. Then the mth power map does the magic by killing all the elements of the group of order n and only permuting the elements of the group of order m. I hope it helps. 

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