math 435 homework5 - 1 ) 1 . . . x ( n + 1 ) n + 1 , ( i )...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 5 for MATH 435 Solutions Problem 1 Book, p. 169, Exercise 2.66 Problem 2 Book, p. 169, Exercise 2.70 Problem 3 Book, p. 170, Exercise 2.78 Problem 4 Book, p. 190, Exercise 2.98 Solution. Let x , y G . Since G/Z ( G ) is cyclic, there exists g G such that G/Z ( G ) = { g m Z ( G ): m Z . Choose m , n Z and z 1 , z 2 Z ( G ) such that x = g m z 1 and y = g n z 2 . Then, using the fact that z 1 , z 2 commute with every element of G and the laws of exponentiation, xy = g m z 1 g n z 2 = g m g n z 1 z 2 = g n g m z 2 z 1 = g n z 2 g m z 1 = yx . ± Problem 5 Book, p. 190, Exercise 2.100 Solution. If G is generated by a single element, then G is cyclic and so is every subgroup. Now let H 6 G , and suppose G is generated by { x 1 , . . . , x n + 1 } . We claim that G/ < x n + 1 > is generated by n elements, namely the cosets x 1 < x n + 1 > , . . . , x n < x n + 1 > . Let g < x n + 1 > be a coset in G/ < x n + 1 > . Since < x 1 , . . . , x n + 1 > = G , and since G is abelian, we can write g = x π (
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 ) 1 . . . x ( n + 1 ) n + 1 , ( i ) Z , and thus g < x n + 1 > = x ( 1 ) 1 . . . x ( n + 1 ) n + 1 < x n + 1 > = x ( 1 ) 1 . . . x ( n ) n < x n + 1 > . Now consider the set H < x n + 1 > = { h < x n + 1 > : h H } . It is not hard to see that this is a sub-group of G/ < x n + 1 > (using the fact that H is a subgroup of G ). By inductive hypothesis H < x n + 1 > is nitely generated, say by { y 1 , . . . y k } . Given h H , we have that h h < x n + 1 > . Thus h can be written as y ( 1 ) 1 . . . y ( k ) k x m n + 1 , with ( i ) , m Z . Hence H is nitely generated....
View Full Document

This note was uploaded on 10/13/2010 for the course MATH 435 at Pennsylvania State University, University Park.

Ask a homework question - tutors are online