4023f11ps3a

4023f11ps3a - Homework #3 Solutions Due: September 14, 2011...

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Homework #3 Solutions Due: September 14, 2011 1. Show that the multiplicative group Z / 14 is cyclic. Find all of its generators. I Solution. Z / 14 = f 1 ; 3 ; 5 ; 9 ; 11 ; 13 g and the powers of 3 mod 14 are: 3 1 = 3, 3 2 = 9, 3 3 = 13, 3 4 = 11, 3 5 = 5, 3 6 = 1. Thus, the powers of 3 flll up Z / 14 so the group is cyclic with generator 3. Since 3 is one generator, the others are 3 k where k is relatively prime to 6 = j Z / 14 j . Thus, k = 1 or k = 5. Hence the generators are 3 1 = 3 and 3 5 = 5. J 2. In each case determine whether G is a cyclic group. (a) G = Z / 7 (b) G = Z / 12 (c) G = Z / 16 (d) G = Z / 11 I Solution. (a) Z / 7 is cyclic with generator 3 since the powers of 3 mod 7 are 3 1 = 3, 3 2 = 2, 3 3 = 6, 3 4 = 4, 3 5 = 5, 3 6 = 1, which gives all of the element of Z / 7 . (b) Z / 12 = f 1 ; 5 ; 7 ; 11 g . The square of each of these elements is 1, so there is no element of order 4. Hence Z / 12 is not cyclic. (c)
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This document was uploaded on 12/28/2011.

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4023f11ps3a - Homework #3 Solutions Due: September 14, 2011...

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