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# 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 a , b = c , d , we need to show that c , d a , b and a , b c , d . 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 ( 2 3 4 ) , ( 1 3 4 2 ) ( 1 2 ) , ( 1 2 3 4 ) . 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 ) ( 2 3 4 ) , ( 1 3 4 2 ) . 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 = ([ 1 ] , ( 1 2 )) . 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)) H ([0],(2 3)) = ([0],(2 3)), ([1],(1 2 3)), ([2],(2 3)), ([0],(1 2 3)), ([1],(2 3)), ([2],(1 2 3)) 4. By Lagrange’s theorem, | A | = 1 , 7 or 49. But 49 is ruled out because

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