4023f10ps6a

4023f10ps6a - Homework #6 Due: October 18, 2010 1. Find all...

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Unformatted text preview: Homework #6 Due: October 18, 2010 1. Find all generators of the cyclic group G = ( g ) if: (a) o ( g ) = 5 (b) o ( g ) = 10 (c) j G j = 16 (d) j G j = 20 I Solution. Use Theorem 7 from the cyclic group supplement. This says that g k is a generator of the cyclic group G = ( g ) if and only if ( n; k ) = 1 where n = j G j = o ( g ). Thus the generators of the given groups are: (a) f g; g 2 ; g 3 ; g 4 g ; (b) f g; g 3 ; g 7 ; g 9 g ; (c) f g; g 3 ; g 5 ; g 7 ; g 9 ; g 11 ; g 13 ; g 15 g ; (d) f g;g 3 ;g 7 ; g 9 ; g 11 ; g 13 ; g 17 ; g 19 g . J 2. In each case determine whether G is cyclic. Recall that Z / n is the group of invertible congruence classes modulo n under multiplication of congruence classes modulo n . (a) G = Z / 7 (b) G = Z / 12 (c) G = Z / 16 (d) G = Z / 11 I Solution. (a) Z / 7 = (3) since the powers of 3 modulo 7 are: 3, 3 2 = 2, 3 3 = 6, 3 4 = 4, 3 5 = 5, 3 6 = 1. Thus, Z / 7 is cyclic. (b) Z / 12 is not cyclic since Z / 12 = f 1 ; 5 ; 7 ; 11 g and a 2 = 1 for all a 2 Z / 12 , so there cannot be an element of order 4 = j Z / 12 j . (c) j Z / 16 j = 8 so look to see if there is an element of order 8 in this group. The powers of 3 modulo 16 are 3, 3 2 = 9, 3 3 = 11, 3 4 = 1. Thus o (3) = 4 and consequently o (11) = 4, o (9) = 2. The powers of 5 modulo 16 are 5, 5 2 = 9, 5 3 = 13, 5 4 = 1. Thus, o (5) = o (13) = 4. The powers of 7 modulo 16 are 7, 7 2 = 1 and the powers of 15 modulo 16 are 15 and 15 2 = 1. Thus o (7) = o (15) = 2. We have accounted for all elements of Z / 16 and none has order 8, so the group is not cyclic....
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4023f10ps6a - Homework #6 Due: October 18, 2010 1. Find all...

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