{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW6_Solutions

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

This preview shows pages 1–2. Sign up to view the full content.

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).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online