math104b-Supplement1

math104b-Supplement1 - 1 Characters. Let G be a commutative...

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1 Characters. Let G be a commutative group with identity element e . Then we recall that a character is a group homomorphism : G ! C = C & f 0 g : That is ( ab ) = ( a ) ( b ) for a; b 2 G . We denote by b G the set of characters of G 1) ( e ) = 1 : This is because ( e ) = ( ee ) = ( e ) 2 : Thus ( e )( ( e ) 1) = 0 : Since ( e ) 6 = 0 , this implies that ( e ) = 1 . 2) ( g ± 1 ) = ( g ) ± 1 . To see this we note that 1 = ( e ) = ( gg ± 1 ) = ( g ) ( g ± 1 ) . 3) If G ( g ± 1 ) = ( g ) for all g 2 G . For this we note that there exists l > 0 such that g l = e . This implies that ( g ) l = 1 . Set z = ( g ) : Since z 6 = 0 there exists w such that z = e w . Thus e lw = 1 and this implies that lw = 2 ±ik with k 2 Z . Thus w = 2 ±i k l . This implies that z = cos(2 ± k l ) + i sin(2 ± k l ) . Also, ( g ± 1 ) = ( g ) ± 1 = e ± w = e ± 2 k l = cos(2 ± k l ) i sin(2 ± k l ) since cos( x ) = cos( x ) and sin( x ) = sin( x ) : 4) We note that the proof of 3) also shows that if G g 2 G then ( g ) l ( g ) = 1 for some l ( g ) > 1 . Lemma 1
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math104b-Supplement1 - 1 Characters. Let G be a commutative...

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