{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# feb07 - Math 4124 Monday February 7 February 7 Ungraded...

This preview shows pages 1–2. Sign up to view the full content.

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 , 0 ) , 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 , 0 ) and ( qa , qb ) = ( 0 , 1 ) . The we have pa = 1 and qb = 1, so p , q = 0, and then pb = 0 and qa = 0 yields a = b = 0 and we have a contradiction. Exercise 2.3.15 on page 60 Prove that Q × Q is not cyclic. Z × Z is a subgroup of Q × Q . From the previous problem Z × Z is not cyclic, which proves the result because subgroups of cyclic groups are cyclic.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

feb07 - Math 4124 Monday February 7 February 7 Ungraded...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online