oct06 - 2 a 3 b where a = , 1 and b = , 1 , 2 , 3, which is...

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Math 3124 Thursday, October 6 October 6 Ungraded Homework Problem 17.1 page 92 Find [ S 3 : h ( 1 2 ) i ] . [ S 3 : h ( 1 2 ) i ] = | S 3 | / 2 = 6 / 2 = 3. Problem 17.3 page 92 Find [ Z 10 : h [ 2 ] i ] . [ Z 10 : h [ 2 ] i ] = | Z 10 | / 5 = 10 / 5 = 2. Problem 17.7 page 92 A group G has subgroups of orders 4 and 10, and | G | < 50. What can you conclude about | G | ? By Lagrange’s theorem, we know that | G | must be a multiple of 4 and 10. Also it is given that | G | < 50. By going through all the possibilities, we see that | G | = 20 or 40. Find the order of ([ 2 ] , [ 2 ]) in Z 10 × Z # 31 . [ 2 ]+[ 2 ]+[ 2 ]+[ 2 ]+[ 2 ] = [ 0 ] in Z 10 , but no smaller sum of [2]’s gives [0]. Therefore [2] has order 5 in Z 10 . [ 2 ] 5 = [ 32 ] = [ 1 ] in Z # 31 , but no smaller power of [2] gives [1]. Therefore [2] has order 5 in Z # 31 . Recall that in G × H , ( g , h ) has order the lowest common multiple of the order of g and the order of h . Therefore ([ 2 ] , [ 2 ]) has order [ 5 , 5 ] = 5 in Z 10 × Z # 31 . The subgroup structure of Z 54 Since Z 54 is a cyclic group of order 54, we have exactly one subgroup for each integer dividing 54. Now 54 = 2 * 3 3 , so the subgroups have order
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Unformatted text preview: 2 a 3 b where a = , 1 and b = , 1 , 2 , 3, which is 2 * 4 possibilities. Therefore Z 54 has 8 subgroups; the orders will be 1,3,9,27 and 2,6,18,54. It follows that the subgroup lattice looks like @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ u h [ ] i u h [ 18 ] i u h [ 27 ] i u h [ 9 ] i u h [ 3 ] i u h [ 1 ] i u h [ 6 ] i u h [ 2 ] i The number of subgroups of a cyclic group of order p 2 q 2 Here p , q are distinct primes. There is exactly one subgroup for each integer dividing the order of the group. The numbers dividing p 2 q 2 are p a q b where a = , 1 , 2 and b = , 1 , 2. Therefore the number of subgroups is 3 * 3 = 9....
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oct06 - 2 a 3 b where a = , 1 and b = , 1 , 2 , 3, which is...

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