extra-homework1

extra-homework1 - & m if x = 0 if x 6 = 0 : If...

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Let G b G to be the set of all group homomorphisms from G as a group under addition to C = C & f 0 g . Such b G as follows: If 1 2 2 b G then ( 1 2 )( x ) = 1 ( x ) 2 ( x ) : The identity in the group is 1 which is our notation for the function ( x ) = 1 for all x 2 G . If G = R m = Z =m Z ; for m > 0 then every character of R m is of the form j ( x ) = e 2 m : With j = 0 ; 1 ; :::; m 1 and x is taken as an element of f 0 ; 1 ; :::; m 1 g . We note that j depends only on its equivalence class modulo m . If y 2 R m ± y : c R m ! C by ± y ( ) = ( y ) . 1) Prove that ± y c R m . 2) Prove that if ± is a character of c R m then there exists y 2 R m such that ± = ± y . 3) Prove that the mapping : R m ! c R m is a group isomorphism (here R m is considered to be a group under addition). 4) Prove that if 2 c R m then X x 2 R m ( x ) = m if = 1 0 if 6 = 1 : (Hint: We may assume that = j for some j .) 5) Prove that if x 2 R m then X 2 d R m ( x ) =
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Unformatted text preview: & m if x = 0 if x 6 = 0 : If f : R m ! C de&ne F ( f ) : c R m ! C by F ( f )( & ) = 1 p m X y 2 R m & ( y ) 1 f ( y ) : 1 6) Prove that f ( x ) = 1 p m X & 2 d R m & ( x ) F ( f )( & ) for all f : R m ! C and all x 2 R m . (Hint: Write out the right hand sum as a double sum and show that the double sum can be written 1 m X & 2 d R m X y 2 R m & ( x & y ) f ( y ) : Make the change of variables z = x & y so y = x & z and write the sum 1 m X z 2 R m ( X & 2 d R m & ( z )) f ( x & z ) : Now use the results above.) If f : R m ! C then we can de&ne a new function b f ( j ) = F f ( & j ) mapping R m to C . 7) Write out the formula for b f ( j ) . 8) Prove that b b f ( x ) = f ( & x ) . 2...
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extra-homework1 - & m if x = 0 if x 6 = 0 : If...

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