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Unformatted text preview: Compute the left cosets of h ( 1 2 ) ih [ 1 ] i in S 3 Z 2 . S 3 Z 2 is a group of order 6 * 2 = 12, while h ( 1 2 ) ih [ 1 ] i is a subgroup of order 2 * 2 = 4, so we expect 12 / 4 = 3 left cosets each with 4 elements. Let H = h ( 1 2 ) ih [ 1 ] i . Then the left cosets of H are H = ((1),[0]), ((1 2),[0]), ((1),[1]), ((1 2),[1]) ((1 2 3),[0])H = ((1 2 3),[0]), ((1 3),[0]), ((1 2 3),[1]), ((1 3),[1]) ((2 3),[0])H = ((2 3),[0]), ((1 3 2),[0]), ((2 3),[1]), ((1 3 2),[1])...
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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
 Algebra, Sets

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