Math 4124 Monday, March 21 March 21, Ungraded Homework Prove that a group of order 175 is abelian. Let G be the group of order 175 = 7 · 5 2 . The number of Sylow 7-subgroups is congruent to 1 mod 7 and divides 25. The only possibility is 1, so G has a normal Sylow 7-subgroup, call it A . Also the number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 7. Therefore G has a normal Sylow 5-subgroup, call it B . Since A and B are of prime or prime squared order, they are both abelian, in particular A ≤ C G ( A ) . Furthermore by Lagrange’s theorem A ∩ B = 1. Since A , B ± G , we deduce that every element of A commutes with every element of B (recall that one proves this by considering aba-1 b-1 = ( aba-1 ) b-1 ∈ B and similarly ∈ A , so aba-1 b-1 ∈ A ∩ B = 1). Thus B ≤ C G ( A ) . By Lagrange’s theorem | A | , | B | divide | C G ( A ) | , so | C G ( A ) | = 175, consequently C G ( A ) = G . This means that A ≤ Z ( G ) . Similarly
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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.