Masters Examination in Mathematics
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Solution: For the action mentioned in the hint, the stabilizer of the coset eH (e is the identity) is H since gH = H implies that g ∈ H. The kernel K of the homomorphism of G into the permutation group of the cosets is a normal subgroup of G which is the intersection of stabilizers of all cosets. In particular the kernel K of the above homomorphism is a subgroup of H. Its index in G is equal to the order of the subgroup of permutation group of cosets which is the image of the above homomorphism and hence has order less that n!.
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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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The details of the inception, development, and implementation for an online masters of mathematics degree, primarily for K-12 mathematics teachers are given. While the program is not yet complete, insofar as all of the courses having being written and taught, we know at this point enough information to begin a process of reflection and refinement about the creation, administration, and operatio...
متن کاملMasters Examination in Mathematics
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...
متن کاملMasters Examination in Mathematics
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...
متن کاملMasters Examination in Mathematics
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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