sep02 - subgroup. The number of Sylow 5-subgroups is...

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Math 5125 Friday, September 2 September 2, Ungraded Homework Exercise 4.5.2 on page 146 Prove that if H is a subgroup of G and Q Syl p ( H ) , then gQg - 1 Syl p ( gHg - 1 ) for all g G . Note that gHg - 1 , gQg - 1 G , so certainly gQg - 1 gHg - 1 . Write | H | = p e m where p - m . Then | Q | = p e , | gHg - 1 | = p e m and | gQg - 1 | = p e . This proves that gQg - 1 Syl p ( gHg - 1 ) . Exercise 4.5.18 on page 147 Prove that a group of order 200 has a normal Sylow 5-
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Unformatted text preview: subgroup. The number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 200 / 25 = 8. The only possibility is 1, which tells us that there is a unique Sylow 5-subgroup which must be normal....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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