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Unformatted text preview: ) a1 = a ( A ∩ H ) a1 = A ∩ H . This proves that A ∩ H ± G as required. 3. Let H ≤ G be ﬁnite groups and let θ : G → G 1 be a group homomorphism. Prove that  θ ( G ) : θ ( H )  divides  G : H  . Let K = ker θ . Then K ∩ H is the kernel of θ restricted to H . By the fundamental isomorphism theorem, we have θ ( G ) ∼ = G / K and θ ( H ) ∼ = H / ( H ∩ K ) . The second isomorphism theorem now yields θ ( H ) ∼ = HK / K . Since G is ﬁnite, so is θ ( G ) and we deduce that  θ ( G ) : θ ( H )  =  θ ( G )  /  θ ( H )  =  G / K  /  HK / K  =  G  /  HK  = (  G  /  H  ) / (  HK  /  H  ) . But  HK  /  K  is an integer by Lagrange’s theorem, so  θ ( G ) : θ ( H )  divides  G  /  H  as required....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

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