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# mar21 - Math 4124 Monday March 21 March 21 Ungraded...

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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 B Z ( G ) and we deduce that | A | , | B | divide | Z ( G ) | . It follows that Z ( G ) has order a multiple of 7 · 25, which is only possible if Z ( G ) = G . This means that
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