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# all2 - Math 3124 Thursday October 13 Fifth Homework...

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Math 3124 Thursday, October 13 Fifth Homework Solutions 1. Problem 14.33 on page 81. Prove or give a counterexample. If a group has a subgroup of order n , then it has an element of order n . S 3 is a subgroup of S 3 of order 6. But the elements of S 3 have order 1, 2 or 3, so we have a counterexample. 2. Problem 15.2 on page 84. Prove that 24 , - 36 , 54 = 6 . 24 6 because 24 = 4 * 6, - 36 6 because - 36 = - 6 * 6, and 54 6 because 54 = 9 * 6. Therefore 24 , - 36 , 54 6 . Conversely 6 = 24 + 2 * ( - 36 ) + 54, so 6 24 , - 36 , 54 and we deduce that 6 24 , - 36 , 54 . The required equality follows. 3. Problem 15.21 on page 84. (a) List the elements of S 3 × Z 2 . (b) List the elements of the cyclic subgroup (( 1 2 ) , [ 1 ]) of S 3 × Z 2 . (c) List the elements of the cyclic subgroup (( 1 2 3 ) , [ 1 ]) of S 3 × Z 2 . (a) ((1), [0]) ((2 3), [0]) ((3 1), [0]) ((1 2), [0]) ((1 2 3), [0]) ((1 3 2), [0]) ((1), [1]) ((2 3), [1]) ((3 1), [1]) ((1 2), [1]) ((1 2 3), [1]) ((1 3 2), [1]) (b) ((1), [0]), ((1 2), [1]) (c) ((1), [0]), ((1 2 3), [1]), ((1 3 2), [0]), ((1), [1]), ((1 2 3), [0]), ((1 3 2), [1]) 4. Problem 16.2 on page 87. Determine the right cosets of [ 3 ] in Z 12 . { [0], [3], [6], [9] } { [1], [4], [7], [10] } { [2], [5], [8], [11] } 5. Problem 16.17 on page 88. Compute the right cosets of (( 1 2 ) , [ 1 ]) in S 3 × Z 2 . Let H = (( 1 2 ) , [ 1 ]) . Then H = { (( 1 ) , [ 0 ]) , (( 1 2 ) , [ 1 ]) } , a subgroup of order 2. Since the order of S 3 × Z 2 is 6 * 2 = 12, we expect 12/2 = 6 cosets. The cosets are H (( 1 ) , [ 0 ]) = { (( 1 ) , [ 0 ]) , (( 1 2 ) , [ 1 ]) } H (( 2 3 ) , [ 0 ]) = { (( 2 3 ) , [ 0 ]) , (( 1 2 3 ) , [ 1 ]) } H (( 3 1 ) , [ 0 ]) = { (( 3 1 ) , [ 0 ]) , (( 1 3 2 ) , [ 1 ]) } H (( 1 2 ) , [ 0 ]) = { (( 1 2 ) , [ 0 ]) , (( 1 ) , [ 1 ]) } H (( 1 2 3 ) , [ 0 ]) = { (( 1 2 3 ) , [ 0 ]) , (( 2 3 ) , [ 1 ]) } H (( 1 3 2 ) , [ 0 ]) = { (( 1 3 2 ) , [ 0 ]) , (( 1 3 ) , [ 1 ]) }

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Math 3124 Thursday, October 20 Sixth Homework Solutions 1. Problem 17.9. Assume that G is a group with a subgroup H such that | H | = 6, [ G : H ] > 4, and | G | < 50. What are the possibilities for | G | ? By Lagrange’s theorem and using the given hypotheses, | G | = 6 [ G : H ] < 50 and [ G : H ] > 4. Therefore the possibilities for [ G : H ] are 5,6,7,8 and hence the possibilities for | G | are 30, 36, 42 and 48. 2. Problem 17.17. Find all the subgroups of Z 36 . Also construct the subgroup lattice. Since Z 36 is a cyclic group of order 36, there will be exactly one subgroup for each positive integer dividing 36. The integers dividing 36 are 2 a 3 b where a , b = 0 , 1 , 2, so there will be 3 * 3 = 9 subgroups. The subgroup lattice will look like [ 0 ] [ 12 ] [ 18 ] [ 6 ] [ 2 ] [ 9 ] [ 4 ] [ 3 ] [ 1 ] 3. From ae = c , we see that neither a nor e is the identity, and from bd = c , we see that neither b nor d is the identity. Therefore c is the identity. Since c is the identity, ae = c yields ea = c , and bd = c yields db = c . We now have the following partial Cayley table a b c d e a a c b b c c a b c d e d c d e c e There are now two ways to complete the table. The first possibility is that da = b . Then the table will be
a b c d e a d e a b c b e a b c d c a b c d e d b c d e a e c d e a b The second possibility is that da = e . Then the table will be a b c d e a b d a e c b d e b c d c a b c d e d e c d a b e c a e b d Remark By considering Z 5 , it can be shown that both the possibilities occur. 4. 22*28 = 616. (a) 616 = 9 * 65 + 31 , 65 = 2 * 31 + 3 , 31 = 10 * 3 + 1 so working from bottom to top, 1 = 31 - 10 * 3 = 31 - 10 * ( 65 - 2 * 31 ) = 21 * 31 - 10 * 65 = 21 * ( 616 - 9 * 65 ) - 10 * 65 = 21 * 616 - 199 * 65 and we deduce that 417 * 65 1 mod 616. Therefore Peter Linnell’s public key is ( 417 , 667 ) .

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all2 - Math 3124 Thursday October 13 Fifth Homework...

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