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Unformatted text preview: Homework 13: Math 3360 Applicable Algebra Chapter 19: Cyclic Groups and Cryptography 15. Show that the group U 10 of units of Z / 10 Z is cyclic. Find a generator of U 1 0. Solution: Here we have U 10 = { 1 , 3 , 7 , 9 } . Clearly h 1 i = { 1 } , h 3 i = { 3 , 3 2 , 3 3 , 3 4 } = { 3 , 9 , 7 , 1 } , h 7 i = { 7 , 9 , 3 , 1 } , and h 9 h{ 9 , 1 } . Thus we see that h 3 i = h 7 h = U 10 , so U 10 is cyclic and is generated by both 3 and 7. 22. If p is prime, then U p is a cyclic group because there is a primitive root modulo p . Show that U p has φ ( p 1) primitive roots. Solution: Let b be a primitive root of U p , so that U p = { b,b 2 ,...,b p 1 = 1 } . It suffices to show that b k is a primitive root, iff ( k,p 1) = 1, because φ ( p 1) is counting the integers smaller than p 1 which are coprime with it. (Note: showing that b k has order ( p 1) / ( k,p 1) is also sufficient, but I didn’t give full credit unless you proved this or cited this result in a theorem). Suppose that ( k,p 1) = d 6 = 1. Then ( b k ) ( p 1) /d = ( b p 1 ) k/d = 1 k/d = 1. Note that this argument works because both ( p 1) /d and k/d are integers. So we see that b k has order less than p 1 and hence is not a primitive root....
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 Spring '08
 BILLERA
 Algebra, Cryptography, Cyclic group, x+1, Primitive root modulo n, primitive root, x+2

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