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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....
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