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Unformatted text preview: Math 4124 Monday, March 28 Solutions to Sample Second Test 1. We have HN / N ∼ = H / H ∩ N . Since we are dealing with finite groups, by taking orders we obtain  HN  /  N  =  H  /  H ∩ N  , hence  HN  /  H  =  N  /  N ∩ H  and we deduce that  HN / H  =  N / N ∩ H  . But  HN / H  divides  G / H  and  N / N ∩ H  divides  N  . Since (  G / H  ,  N  ) = 1, we deduce that  N / N ∩ H  = 1. Therefore N = N ∩ H which shows that N ⊆ H . 2. If g ∈ G , then tg 6 = g . This means when we write out π t as a product of disjoint cycles, it cannot contain any 1cycles; also t 2 g = 1 g = g , so the other cycles are all 2cycles. In other words π t is a product of 12 disjoint 2cycles (each of the cycles will be of the form ( g , tg ) for some g ∈ G ). In particular π t ∈ A G ∼ = A 24 . If g ∈ S 4 , then g is a product of transpositions (not disjoint in general, and certainly not uniquely so) so we may write g = t 1 . . . t n where the t i are transpositions. Then π g = ( π t 1 ) . . . ( π t n ) because π is a homomorphism. From the first paragraph, π t i ∈ A G for all i and we deduce that π g ∈ A G . Finally ker π = 1, so π is a onetoone homomorphism whose image is contained in A G . We deduce that G is isomorphic to a subgroup of A 24 . 3. First we compute the size of the conjugacy class. We have 10 · 9 ····· 3 ways to choose the eight entries. Then we can rotate each cycle; that means we have to divide by 3 · 3 · 2. Finally we can interchange the two 3cycles; that means we need to divide2....
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.
 Spring '08
 Staff
 Math, Algebra

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