Masters Examination in Mathematics
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Let G have order 2013. For p = 3, 11, 61 denote by np the number of Sylow p-groups in G. By Sylow’s theorems we have 61 | (n61 − 1) and n61|33, which is possible only for n61 = 1. Hence the 61-Sylow subgroup B is unique and therefore normal in G. Similarly, n11 | 3× 61 and 11 | (n11 − 1) yields n11 = 1; and the unique 11-Sylow subgroup A is normal in G. Note that A ∩ B is the trivial subgroup {e} because its order has to divide both 11 and 61. At the same time every a ∈ A commutes with every b ∈ B because
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Masters Examination in Mathematics
Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth 20 points. All answers will be graded, but the score for the examination will be the sum of the scores of your best 8 solutions. Algebra A1. (a) State the Class Equation.
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Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth 20 points. All answers will be graded, but the score for the examination will be the sum of the scores of your best 8 solutions. Use separate answer sheets for each question. DO NOT PUT YOUR NAME ON YOUR ANSWER SHEETS. When you have finished, insert all your answer shee...
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Solution: The ring F [x] of polynomials with coefficients in a field F is a P.I.D. Each prime ideal is generated by a monic, irreducible polynomial. Assume there are only a finite number of prime ideals generated by the polynomials f1, . . . , fn and let f(x) = 1+f1(x) · · · fn(x). No fi divides f , hence f is also irreducible. This contradicts the assumption that all the prime ideals were gene...
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Solution. The 11-Sylow subgroup is Z/11Z; the 5-Sylow subgroup is Z/5Z. By Sylow’s theorem, the 11-Sylow subgroup is normal. Hence, the group is a semi-direct product of its 5 and 11-Sylow subgroups. Since Aut(Z/11Z) = Z/10Z has a unique subgroup of order 5, there are up to isomorphism exactly two groups of order 55: the abelian group Z/55Z and the group with presentation ⟨x, y|x = y = 1, xyx−1...
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Solution. Since (ab)(abc) = (bc) and (abc)(ab) = (ac), it is easy to see that the center of S3 is the trivial subgroup. Therefore, the group of inner automorphisms of S3 is isomorphic to S3 and has size 6. On the other hand, S3 has 3 transpositions. These must be permuted by an automorphism of S3, and a nontrivial automorphism induces a nontrivial permutation. This gives an injective homomorphi...
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