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Unformatted text preview: 18.781, Fall 2007 Problem Set 5 Solutions to Selected Problems Problem 2.11.1 By the argument of from 122 page to 123 page, since 9 = 3 2 , the multiplicative group modulo 9 is a cyclic group of order (9) = 6. It is clear that the additive group modulo 6 is a cyclic group of order 6. Hence, their isomorphic. More precisely, define a function f : R 9 Z 6 by f (2) = 1 ,f (4) = 2 ,f (8) = 3 ,f (7) = 4 ,f (5) = 5 ,f (1) = 0 . ( For each a R 9 , there is an g such that 2 g a in modulo 9 because 2 is a primitive root of 9, and we define f ( a ) = g where g lies in modulo 6. This is well defined since 6 is the order of 2 in modulo 9. Then f is bijective, clearly. To show that f is an isomorphism, we need to show that f is a group homomorphism. For a,b R 9 , there are g,h such that 2 g a and 2 h b in modulo 9. So we have f ( ab ) f (2 g 2 h ) = f (2 g + h ) g + h f (2 g ) + f (2 h ) f ( a ) + f ( b ) (mod 6). Therefore f is a group homomorphism. 2 Problem 2.11.6 By the argument from 122 page to 123 page, R m is a cyclic group if and only if m = 2 , 4 ,p,p . For 2 , 3 , 4 , 5 , 6 , 7, each of them is one of the previous form, and 8 is not. Therefore, 8 is the smallest positive integer m such that the multiplicative group modulo m is not cyclic. 2 Problem 2.11.11 Solution 1 If we proved that G is a group, it is clear that G is noncommutative because a b = d 6 = f = b a . Looking at the table, we can easily find that e is an identity, and each element of a,b,c,d,f has an inverse element ( a a = b b = c c = e,d f = f d = e ). It remains to prove the associativity. We need to show that x ( y z ) = ( x y ) z for x,y,z { e,a,b,c,d,f } . If one of x,y,z is e , the associativity clearly holds. For the other cases, For ( x,y,z ) = ( a,a,a ), we have x ( y z ) = a = ( x y ) z . For ( x,y,z ) = ( a,a,b ), we have x ( y z ) = b = ( x y ) z . For ( x,y,z ) = ( a,a,c ), we have x ( y z ) = c = ( x y ) z . For ( x,y,z ) = ( a,a,d ), we have x ( y z ) = d = ( x y ) z . 1 For ( x,y,z ) = ( a,a,f ), we have x ( y z ) = f = ( x y ) z . For ( x,y,z ) = ( a,b,a ), we have x ( y z ) = c = ( x y ) z . For ( x,y,z ) = ( a,b,b ), we have x ( y z ) = a = ( x y ) z . For ( x,y,z ) = ( a,b,c ), we have x ( y z ) = b = ( x y ) z . For ( x,y,z ) = ( a,b,d ), we have x ( y z ) = f = ( x y ) z . For ( x,y,z ) = ( a,b,f ), we have x ( y z ) = e = ( x y ) z . ....
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This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.
 Spring '09
 BRUBAKER
 Number Theory

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