Adv Alegbra HW Solutions 67

Adv Alegbra HW Solutions 67 - f (( h , 1 )) , and ker f = {...

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67 where x H . First, we show that g is a homomorphism. If x , y H , then g ( xy ) = ( xy , 1 ) K . On the other hand, the de f nition of multipli- cation in the quotient group ( H × K )/ K gives g ( x ) g ( y ) = ( x , 1 ) K ( y , 1 ) K = ( xy , 1 ) K = g ( xy ), as desired. Next, we show that g is an injection. Now if x H ,wehave g ( x ) = ( x , 1 ) K ={ ( x , 1 )( 1 , k ) : k K }={ ( x , k ) : k K } . It follows that if g ( x ) = g ( y ) , then { ( x , k ) : k K }={ ( y , k ) : k K } , and this implies that x = y . Finally, we show that g is a surjection. If ( x , k ) K ( H × K )/ K , then ( x , k ) K = ( x , 1 ) K [recall Lemma 2.82(i): aH = bH if and only if b 1 a H ; here, ( x , 1 ) 1 ( x , k ) = ( 1 , k ) K ]. Therefore, ( x , k ) K = ( x , 1 ) K = g ( x ) , and so g is surjective. Therefore, g is an isomorphism. (iii) Prove K C ( H × K ) and ( H × K )/ K = H using the f rst isomorphism theorem. Solution. The function f : H × K H ,de f ned by f : ( h , k ) 7→ h , is a homomorphism: f (( h , k )( h 0 , k 0 )) = f (( hh 0 , kk 0 )) = hh 0 = f (( h , k )) f (( h 0 , k 0 )). Now f is surjective, for if h H , then h =
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Unformatted text preview: f (( h , 1 )) , and ker f = { ( h , k ) H K : f (( h , k )) = 1 } = { ( 1 , k ) H K : k K } = K . The f rst isomorphism theorem now gives K C H K and ( H K )/ K = H . (Of course, the function H K K , sending ( h , k ) 7 k , is a homomorphism with kernel H , and this implies that H C H K . 2.98 If G is a group and G / Z ( G ) is cyclic, where Z ( G ) denotes the center of G , prove that G is abelian; that is, G = Z ( G ) . Conclude that if G is not abelian, then G / Z ( G ) is never cyclic....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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