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December 3rd, 2012, 03:56 PM   #1
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Order of a Subgroup M


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!
flostamjets is offline  
December 24th, 2012, 01:48 AM   #2
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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 p-group 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 m-th 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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