mar23 - Math 4124 Wednesday March 23 March 23 Ungraded...

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Math 4124 Wednesday, March 23 March 23, Ungraded Homework Exercise 4.5.18 on page 147. Prove that a group of order 200 has a normal Sylow 5- subgroup. 200 = 8*25. Therefore the number of Sylow 5-subgroups divides 8 and is congruent to 1 mod 5. The only possibility is 1, which shows that the Sylow 5-subgroup is normal. Exercise 4.5.26 on page 147. Let G be a group of order 105. Prove that if a Sylow 3- subgroup of G is normal in G , then G is abelian. 105 = 3*5*7. We want to show that the Sylow 5 and 7-subgroups are abelian, and then it will quickly follow that G is abelian. Now we cannot immediately assert that the number of Sylow 7-subgroups is 1, because it appears at first glance that 15 is a possibility (15 is congruent to 1 mod 7 and divides 3*5). Let A be a Sylow 3-subgroup of G . We know that A ± G , so we may consider G / A , a group of order 105 / 3 = 35. The number of Sylow 7- subgroups of G / A is congruent to 1 mod 7 and divides 5, hence G / A has exactly on Sylow 7-subgroup and we deduce that G / A has a normal Sylow 7-subgroup. By the subgroup correspondence theorem, we may write this subgroup as P / A , where P ± G and | P | = 7 | A | = 21. The number of Sylow 7-subgroups of P is congruent to 1 mod 7 and divides 3, hence this number is 1 and we see that P has a normal Sylow 7-subgroup, which we shall call B . Though the property of being normal is not in general transitive, it is true that
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.

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mar23 - Math 4124 Wednesday March 23 March 23 Ungraded...

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