Partial Solutions to Homework 09

# Partial Solutions to Homework 09 - MATH 74 HOMEWORK 9 Due...

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MATH 74 HOMEWORK 9 Due Wednesday April 24th at 3:10pm. Throughout, let G be a group. Many of the following problems are from (or adapted from) Herestein’s Topics in Algebra . 1 Let Z ( G ) be the center of G and suppose that G/Z ( G ) is cyclic. Prove that G is abelian. Solution. Since Z ( G ) is a subgroup, G can be written as the disjoint union of the cosets of Z ( G ) in G . Since the quotient is cyclic (say with generator g ) these cosets are precisely Z ( G ) g k for k and integer. Then any elements a,b can be written as a = z 1 g n , b = z 2 g m for z 1 ,z 2 Z ( G ) and m,n integers. Note that z 1 and z 2 commute with all elements of G by virtue of being in the center of G . So we compute ab = z 1 g n z 2 g m = z 1 z 2 g m + n = z 2 z 1 g m g n = z 2 g m z 1 g n = ba, so G is abelian. 2 Prove that any group G of order 15 is cyclic. Solution. By Cauchy’s theorem, there are elements x and y of orders 3 and 5 respectively. If x and y commute with each other, we obtain a homomorphism, Z / 3 Z × Z / 5 Z G, ( i,j ) 7→ x i y j with trivial kernel. Then

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## This note was uploaded on 05/04/2011 for the course MATH 73 taught by Professor Danberwick during the Spring '09 term at Berkeley.

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Partial Solutions to Homework 09 - MATH 74 HOMEWORK 9 Due...

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