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Adv Alegbra HW Solutions 67

Adv Alegbra HW Solutions 67 - f h 1 and ker f = h k ∈ H...

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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 fi 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 , we have 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 ( H × K ) and ( H × K )/ K = H using the fi rst isomorphism theorem. Solution. The function f : H × K H , de fi ned by f : ( h , k ) h , is a homomorphism: f (( h , k )( h , k )) = f (( hh , kk )) = hh = f (( h , k )) f (( h , k )). Now f is surjective, for if 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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