This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 4124 Monday, February 7 February 7, Ungraded Homework Exercise 2.3.12 on page 60 Prove that the following groups are not cyclic. (a) Z / 2 Z Z / 2 Z (b) Z / 2 Z Z (c) Z Z (a)  Z / 2 Z Z / 2 Z  = 4, so if the group was cyclic, it would have an element of order 4. This is not the case because all nonidentity elements have order 2. (b) Since Z / 2 Z Z is infinite, if it is also cyclic, it must be isomorphic to Z . However the given group has an element of order 2, namely ( 1 , ) , whereas all nonidentity elements of Z have infinite order. (c) Suppose Z Z is cyclic. Then it would have a generator ( a , b ) , where a , b Z . This would mean that every element of the group could be written as ( na , nb ) for some n Z . Thus there would be p , q Z such that ( pa , pb ) = ( 1 , ) and ( qa , qb ) = ( , 1 ) . The we have pa = 1 and qb = 1, so p , q 6 = 0, and then pb = 0 and qa = 0 yields a = b = 0 and we have a contradiction....
View
Full
Document
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.
 Spring '08
 Staff
 Math, Algebra

Click to edit the document details