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November 3rd, 2011, 06:05 PM  #1 
Newbie Joined: Oct 2011 Posts: 2 Thanks: 0  Herstein: Abstract Algebra Proof (Sylow's Theorem) if P^m divides G, show that G has a subgroup of order P^m 
November 11th, 2011, 04:25 AM  #2 
Member Joined: Jun 2010 Posts: 64 Thanks: 0  Re: Herstein: Abstract Algebra Proof (Sylow's Theorem)
We have to solve it by sylow theorem, P^m divides G then we have that pG,so exist an integer n such that P^n G and n is the greatest integer that P^nG, By the sylow theorem exist a subgroup H<G with order P^n. so H is pgroup (p is simple) We have that m<=n; you have to find a subgroup D<H<G that the order of D=p^m . The existence of the group D is guaranted by a theorem in pgroup. Use this theorem below Theorem: if G(in your case H) is a group of order p^n, then exist a subgroup D of G with order p^i , for i=1,2,,,,,m1,m This theorem guaranted the existence of a subgroup H with order p^m (m<=n) Note : In your question you have to add that pis prime,.becouse in general is not true that "for p not prime,and P^m divides G, show that G has a subgroup of order P^m , ," 

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