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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 Fall '11
 KeithCornell
 Cyclic group, Homomorphism, kernel, Epimorphism

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