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November 3rd, 2011, 07:05 PM   #1
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Herstein: Abstract Algebra Proof (Sylow's Theorem)

if P^m divides |G|, show that G has a subgroup of order P^m
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November 11th, 2011, 05:25 AM   #2
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Re: Herstein: Abstract Algebra Proof (Sylow's Theorem)

We have to solve it by sylow theorem,
P^m divides |G| then we have that p||G|,so exist an integer n such that P^n| |G| and n is the greatest
integer that P^n||G|, By the sylow theorem exist a subgroup H<G with order P^n.
so H is p-group (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 p-group.
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,,,,,m-1,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 p-is 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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