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Topic: Lagrange's Theorem (Read 845 times) 

Obob
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Re: Lagrange's Theorem
« Reply #25 on: Sep 22^{nd}, 2007, 7:24pm » 
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Yes, Eigenray, you are right. I was trying to make the proof as simple as possible, and ended up making it simpler than possible . For a real proof, note that any group of order p^{a} has nontrivial center. This follows from letting G act on itself by conjugation; then all orbits have order a power of p, and since their sum is the order of G and the identity has an orbit of size 1, there must be other orbits of size 1. Now pick a nonidentity element g from the center. Some power h of g has order p, and <h> is normal. So we can consider G/<h>, and proceed by induction.

« Last Edit: Sep 22^{nd}, 2007, 7:24pm by Obob » 
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ecoist
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Re: Lagrange's Theorem
« Reply #26 on: Oct 8^{th}, 2007, 2:29pm » 
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Solution for Eigenray's easier problem: Determine those integers n such that if n divides the order of a group G, then G has a subgroup of index n. We show that the only such n is n=1. Let n>1. Let G be the alternating group on 2n letters. Then G is a simple group of order greater than n!, and its order is divisible by n. If G had a subgroup of index n, then G would be isomorphic to a subgroup of the symmetric group on n letters, contradicting the fact that G has order larger than n!.


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