ahw7 - Math 3124 Thursday October 27 Seventh Homework...

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Math 3124 Thursday, October 27 Seventh Homework Solutions 1. S 5 × S 4 has elements of order 15, for example ((1 2 3 4 5), (1 2 3)). On the other there is no element of order 15 in S 6 (the only way to have a permutation of order 15 is if there is a 15 cycle, or a product of a 5 cycle and a 3 cycle (which are disjoint), which would require at least S 8 ). Since the elements of Z 4 have order 1, 2 or 4 and o ( x , y ) = [ o ( x ) , o ( y )] , we see that S 6 × Z 4 can have an element of order 15 only if S 6 has an element of order 15. Therefore S 6 × Z 4 has no element of order 15 and it follows that S 5 × S 4 ± S 6 × Z 4 . 2. Problem 21.6(a) on page 109. Show that α : Z 3 Z 6 by α ([ a ] 3 ) = [ a ] 6 is not well- defined. [ 0 ] 3 = [ 3 ] 3 , so for α to be well defined, we must have α [ 0 ] 3 = α [ 3 ] 3 , that is [ 0 ] 6 = [ 3 ] 6 . But [ 0 ] 6 6 = [ 3 ] 6 and it follows that α is not well-defined. 3. Problem 21.27 on page 110. Prove that if
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ahw7 - Math 3124 Thursday October 27 Seventh Homework...

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