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# HW5_soln - Homework#5 Solutions p 83#16 In order to nd a...

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Homework #5 Solutions p 83, #16. In order to find a chain a 1 a 2 ≤ · · · ≤ a n of subgroups of Z 240 with n as large as possible, we start at the top with a n = 1 so that a n = Z 240 . In general, given a i we will choose a i - 1 to be the largest proper subgroup of a i . We will make repeated use of the fundamental theorem of cyclic groups which tells us that a cyclic group of order m has a unique subgroup of order d for any d | m . The largest proper subgroup of Z 240 has size 120 and is 2 . Since | 2 | = 120, the largest proper subgroup of 2 has size 60 and is 4 . Since | 4 | = 60, the largest proper subgroup of 4 has size 30 and is 8 . Since | 8 | = 30, the largest proper subgroup of 8 has order 15 and is 16 . Since | 16 | = 15, the largest possible subgroup of 16 has order 5 and is 48. Finally, since | 48 | = 5 is prime, the only proper subgroup of 48 is 0 . Therefore, we have produced the maximal chain 0 48 16 8 4 2 1 which has length 7. Notice that the chain 0 120 60 30 15 5 1 also has length 7, but is produced in the opposite way, i.e. by starting with 0 and at each stage choosing a i +1 as the smallest subgroup containing a i . p 83, # 20. Let x G . Since x 35 = e , we know that | x | = 1 , 5 , 7 or 35. Since | G | = 35, if G contains an element x of order 35, then G = x as desired. On the other hand, if G contains an element x of order 5 and and element y of order 7, then, since G is abelian ( xy ) 35 = x 35 y 35 = ee = e so that the order k of xy divides 35. That is, | xy | = 5 , 7 or 35. If | xy | = 5 then e = ( xy ) 5 = x 5 y 5 = ey 5 = y 5 which means that 7 = | y | divides 5, a contradiction. Likewise, we have a similar problem if | xy | = 7. It follows that | xy | = 35, and as above that G is cyclic. So, what we need to do is show that G must have an element of order 5 and an element of order 7. We argue by contradiction. If G has no elements of order 5 then every non-identity element of G has order 7. That is, there are 34 elements in G or order 7. However, by the corollary to Theorem 4.4, the number of elements in G of order 7 is divisible by φ (7) = 6, and 34 is not divisible by 6. Likewise, if G had no element of order 7 then G would contain 34 elements of order 5, and this number would have to be divisible by φ (5) = 4, which is also impossible. It follows that G must have at least one element of order 5 and at least one of order 7. As we pointed out above, this forces G to be cyclic.

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HW5_soln - Homework#5 Solutions p 83#16 In order to nd a...

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