midterm1.soln.f08

midterm1.soln.f08 - Math330 Solutions to Midterm 1 Fall...

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Math330 Solutions to Midterm 1 Fall 2008 1. The set { 1 , 2 , 3 } is not a group under multiplication modulo 4 because it is not closed. In particular, 2 × 2 = 4 so 2 · 4 2 = 0 . Therefore, { 1 , 2 , 3 } cannot be a group. One could also note that 1 is the identity and 2 does not have an inverse (since 2 · 4 1 = 2, 2 · 4 2 = 0 and 2 · 4 3 = 2). 2. The group U (13) = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } is a group under multiplication modulo 13. 1. Given that U (13) is a cyclic group of order 12, we can determine how many gener- ators the group has by taking φ (12) = 4. (Corresponding to 1 , 5 , 7 , 11; the numbers relatively prime to 12.) The cyclic group U (13) has 4 generators. 2. Given that the number 2 is a generator of U (13), we know that | 2 | = | U (13) | = 12 . 3. To calculate the order of the element 2 8 in U (13) we use Theorem 4.2 on page 76. Thus, | 2 8 | = 12 / gcd(12 , 8) = 12 / 4 = 3 . 4. The number of generators of
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midterm1.soln.f08 - Math330 Solutions to Midterm 1 Fall...

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