HW6_Solutions

HW6_Solutions - Math 541 Solutions to HW #6 The following...

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Unformatted text preview: Math 541 Solutions to HW #6 The following are from Gallian, Chapters 4 and 5 (6th edition). # 4.8 : Let a be an element of a group and let | a | = 15. Compute the orders of the following elements of G : a 3 , a 6 , a 9 , a 12 * For each a k above, gcd(15 ,k ) = 3. Thus, the order of each is 15 / 3 = 5. a 5 , a 10 * For each a k above, gcd(15 ,k ) = 5. Thus, the order of each is 15 / 5 = 3. a 2 , a 4 , a 8 , a 14 * For each a k above, gcd(15 ,k ) = 1. Thus, the order of each is 15 / 1 = 15. # 4.14 : Suppose that a cyclic group G has exactly three subgroups: G itself, { e } , and a subgroup of order 7. What is | G | ? What can you say if 7 is replaced with p where p is a prime? Since G is cyclic, there is some element a in G such that h a i = G . Since G has a subgroup of order 7, and G is cyclic, we know that 7 divides the order of G . That is, | h a i | = | G | = 7 n for some positive integer n . We now test a few possible values of n : * Suppose n = 1. Then G and one of its subgroups both have order 7. By the Fundamental Theorem of Cyclic Groups (FTCG), G and its subgroup of order 7 are the same, contradicting the condition that G has 3 distinct subgroups. * Suppose n is 2, 3, 4, 5, or 6. Then, by FTCG, G = h a i has a subgroup of order n . Thus, G has at least 4 subgroups: { e } , the subgroup of order 7, the subgroup of order n , and G itself. This contradicts the fact that G has exactly three subgroups. * Suppose n = 7. Then | G | = 7 7 = 49. Since 7 is the only positive divisor of 49 between 1 and 49, it is the only possible order of a subgroup other than { e } or G . FTCG also tells us that there is exactly one subgroup of order 7. This fits the supposed criteria. * In general, if we suppose that n is any positive integer besides 7, we see that G is guaranteed a subgroup of order n by the FTCG, which means that G will have at least 4 distinct subgroups. We therefore conclude that the order of G must be 7 2 = 49. More generally, if 7 is replaced by any prime p under the supposed conditions, the the order of G must be p 2 . # 4.16 : Find a collection of distinct subgroups h a 1 i , h a 2 i , ..., h a n i of Z 240 with the property that h a 1 i h a 2 i ... h a n i with n as large as possible. Since Z 240 is cyclic and the order of a subgroup of a cyclic group divides the order of the group in which it is contained, we see it must be true that | h a i i | divides | h a i +1 i | ,..., | h a n- 1 i | , | h a n i | . That is, the order of a subgroup divides the order of every subgroup in which it is contained. Breaking 240 into its prime factorization, we get 240 = 2 4 3 5. That is, 240 is the product of 6 primes (note that they need not be distinct)....
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This note was uploaded on 11/03/2010 for the course MATH 541 taught by Professor Pollack during the Fall '09 term at BU.

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HW6_Solutions - Math 541 Solutions to HW #6 The following...

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