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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 1-cycles; also t 2 g = 1 g = g , so the other cycles are all 2-cycles. In other words π t is a product of 12 disjoint 2-cycles (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 one-to-one 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 3-cycles; that means we need to divide2....
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