ahw1 - ) a-1 = a ( A ∩ H ) a-1 = A ∩ H . This proves...

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Math 5125 Monday, August 29 First Homework Solutions 1. Section 3.2, Exercise 8 on page 95. Prove that if H and K are finite subgroups of the group G whose orders are relatively prime, then H K = 1. Since H K is a subgroup of the finite group H , we see that | H K | divides | H | by Lagrange’s theorem. Similarly | H K | divides | K | . Since ( | H | , | K | ) = 1, we deduce that | H K | = 1 and we conclude that H K = 1 as required. 2. Let A be a normal abelian subgroup of the group G and let H G . Suppose AH = G . Prove that A H ± G . Hint: if g G , write g = ah and then show a ( A H ) a - 1 = A H = h ( A H ) h - 1 . Certainly A H is a subgroup of G because the intersection of two subgroups is a subgroup, so we need to prove normality, that is g ( A H ) g - 1 = A H for all g G . Since G = AH , we may write g = ah with a A and h H . Now A is abelian and all subgroups in an abelian group are normal, in particular A H ± A and we see that a ( A H ) a - 1 = A H . Further more hAh - 1 = A because A ± G and hHh - 1 = H because h H . Therefore h ( A H ) h - 1 = A H . We deduce that g ( A H ) g - 1 = ah ( A H )( ah ) - 1 = a ( h ( A H ) h - 1
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Unformatted text preview: ) a-1 = a ( A ∩ H ) a-1 = 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.

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