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# sep09 - | H H ∩ P | = | HP P | Also | G P | = | G HP ||...

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Math 5125 Friday, September 9 September 9, Ungraded Homework Exercise 4.5.33 on page 147 Let P be a normal Sylow p -subgroup of G and let H be any subgroup of G . Prove that P H is the unique Sylow p -subgroup of H . First we prove that P H is a Sylow p -subgroup of H . Obviously P H is a p -subgroup of H . To show it is actually a Sylow p -subgroup, we need to show that it is divisible by the highest possible power of p which is consistent with Lagrange’s theorem, equivalently | H : H P | is not divisible by p . By the second isomorphism theorem,
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Unformatted text preview: | H : H ∩ P | = | HP : P | . Also | G : P | = | G : HP || HP : P | and | G : P | is not divisible by p , because P is a Sylow p-subgroup of G . It follows that | H : H ∩ P | is not divisible by p , which establishes that H ∩ P is a Sylow p-subgroup of H . Also it is a normal subgroup of H (for example, by the second isomorphism theorem). By Sylow’s theorems, we conclude that H ∩ P is the unique Sylow p-subgroup of G ....
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