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Unformatted text preview: 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 ) , [ ]) , (( 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 ) , [ ]) = { (( 1 ) , [ ]) , (( 1 2 ) , [ 1 ]) } H (( 2 3 ) , [ ]) = { (( 2 3 ) , [ ]) , (( 1 2 3 ) , [ 1 ]) } H (( 3 1 ) , [ ]) = { (( 3 1 ) , [ ]) , (( 1 3 2 ) , [ 1 ]) } H (( 1 2 ) , [ ]) = { (( 1 2 ) , [ ]) , (( 1 ) , [ 1 ]) } H (( 1 2 3 ) , [ ]) = { (( 1 2 3 ) , [ ]) , (( 2 3 ) , [ 1 ]) } H (( 1 3 2 ) , [ ]) = { (( 1 3 2 ) , [ ]) , (( 1 3 ) , [ 1 ]) } 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 Lagranges 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 = , 1 , 2, so there will be 3 * 3 = 9 subgroups. The subgroup lattice will look like [ ] [ 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....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.
 Fall '08
 PARRY
 Math, Algebra

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