ast2 - Math 3124 Tuesday, October 25 Second Sample Test...

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Math 3124 Tuesday, October 25 Second Sample Test Solutions 1. To show that h a , b i = h c , d i , we need to show that c , d ∈ h a , b i and a , b ∈ h c , d i . Now (2 3 4) = (1 2)(1 2 3 4) and (1 3 4 2) = (2 3 4)(1 2) = (1 2)(1 2 3 4)(1 2). This tells us that h ( 2 3 4 ) , ( 1 3 4 2 ) i ⊆ h ( 1 2 ) , ( 1 2 3 4 ) i . On the other hand (2 3 4)(1 3 4 2) = (1 4 3 2), so by taking inverses, we see that ( 1 2 3 4 ) ∈ h ( 2 3 4 ) , ( 1 3 4 2 ) i . Finally (2 3 4)(1 4 3 2) = (1 2) and the result follows. 2. First note that bc = c forces b to be the identity. Then ad cannot be a , d , b , so must be c etc. a b c d a b a d c b a b c d c d c a b d c d b a 3. The order of Z 3 is 3, and the order of S 3 is 6. Therefore the order of Z 3 × S 3 is 3 * 6 = 18. Let H = h ([ 1 ] , ( 1 2 )) i . The order of [1] in Z 3 is 3 and the order of (1 2) in S 3 is 2. Therefore the order of ([1],(1 2)) is 3 * 2 = 6 and we deduce that | H | = 6. Thus we expect 3 right cosets with 6 elements each. The right cosets are H = ([0],(1)), ([1],(1 2)), ([2],(1)), ([0],(1 2)), ([1],(1)), ([2],(1 2)) H ([0],(1 3)) = ([0],(1 3)), ([1],(1 3 2)), ([2],(1 3)), ([0],(1 3 2)), ([1],(1 3)), ([2],(1 3 2))
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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.

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ast2 - Math 3124 Tuesday, October 25 Second Sample Test...

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