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Unformatted text preview: Algebra , Assignment 2 Solutions Due, Friday, Sept 24 1. In S 3 (reminder, this is our standard notation for the permutations on { 1 , 2 , 3 } ) show that there are four elements x satisfying x 2 = e and three elements x satisfying x 3 = e . Solution: The flips parenleftbigg 123 132 parenrightbigg , parenleftbigg 123 321 parenrightbigg , parenleftbigg 123 213 parenrightbigg have x 2 = e and parenleftbigg 123 231 parenrightbigg and parenleftbigg 123 312 parenrightbigg have y 3 = e and e satisfies both x 2 = e and y 3 = e . 2. In Z * 13 let H = { 1 , 5 , 12 , 8 } . List the right cosets Ha . Solution: H 1 = { 1 , 5 , 12 , 8 } , H 2 = { 2 , 10 , 11 , 3 } , H 4 = { 4 , 7 , 9 , 6 } and that is all of them. (As the group Z * 13 is Abelian right and left cosets are the same.) 3. Let G be the symmetries of the square. (See the solutions to assign ment 1 for a table.) Let H = { I,V } . List the right cosets Ha and the left cosets aH . Do the same with H = { I,R,S,T } ....
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 Fall '08
 JOSEPHS
 Permutations

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