Fa11_withsolutionsPM

# Fa11_withsolutionsPM - MASTERS EXAMINATION IN MATHEMATICS...

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Unformatted text preview: MASTERS EXAMINATION IN MATHEMATICS PURE MATH OPTION, FALL 2011 Algebra A1. Suppose that G is a group, with normal subgroups A and B so that G/A and G/B are abelian. Prove that G/ ( A ∩ B ) is abelian. Solution. A group G/K is abelian if and only if the commutator subgroup [ G,G ] is con- tained in K . If G/A and G/B are abelian then [ G,G ] ≤ A and [ G,G ] ≤ B so [ G,G ] ≤ A ∩ B . Therefore, G/ ( A ∩ B ) is abelian. A2. Prove that if the order of G is 132, then G is not simple. Solution. Let n p denote the number of p-Sylow subgroups. Since n 11 is 1 mod 11 and divides 12, we know n 11 is either 1 or 12. If it is 1, then it is normal and we are done. So instead, assume n 11 is 12. Then there must be 120 elements of order 11 in G , leaving only 12 more elements available. Since n 3 is 1 mod 3 and divides 44, we know that n 3 is 1 , 4 or 22. If it is 1, then again it is normal and we would be done. If it is 22, that would require 44 elements of order 3 which is not possible. If n 3 is 4 that only requires 8 elements of order 3. This leaves 4 elements available which must form the unique Sylow 2-subgroup, which is then normal. A3. Prove that the polynomial x 2- √ 2 is irreducible over Z [ √ 2]. Solution. Suppose not. Since x 2- √ 2 is monic, if it factorizes as a product of two non-units, these non-units must each be linear. Then there are α,β,γ,δ ∈ Z [ √ 2] so that x 2- √ 2 = ( αx + β )( γx + δ ) . Since αγ = 1 we get x 2- √ 2 = ( x + βγ )( x + δα ) . Let βγ = u and δα = v . Then we have uv =- √ 2 and u + v = 0, which means that u 2 = √ 2. However, if u = a + b √ 2 then u 2 = ( a 2 + 2 b 2 ) + (2 ab ) √ 2. Since a,b ∈ Z it is clear that we cannot have u 2 = √ 2. Complex Analysis 1 C1. Does the integral Z ∞-∞ cos( x ) x 2 + 2 x + 2 dx converge? If so, what is its value? Solution. The integral converges since cos( x ) x 2 + 2 x + 2 ≤ 1 x 2 + 2 x + 2 and Z ∞-∞ 1 x 2 + 2 x + 2 dx converges. To compute its value we note that Z ∞-∞ cos( x ) x 2 + 2 x + 2 dx = P .V. Z ∞-∞ cos( x ) x 2 + 2 x + 2 dx = lim R →∞ R e Z R- R e ix x 2 + 2 x + 2 dx = R e 2 πiRes z =- 1+ i e iz z 2 + 2 z + 2- lim R →∞ R e Z C R e iz z 2 + 2 z + 2 dz, where C R is the upper half of the circle | z | = R from z = R to z =- R . Note that Res z =- 1+ i e iz z 2 + 2 z + 2 = e- 1- i 2 i and for R large enough Z C R e iz z 2 + 2 z + 2 dz ≤ πR · 1 R 2- 2 R- 2 → , as R → ∞ . Therefore Z ∞-∞ cos( x ) x 2 + 2 x + 2 dx = R eπe- 2- i = πe- 2 cos 1 . C2. Find all singular points of the function h ( z ) = e 1 z- 3 sin 2 ( z ) ( z- 1) z 2 2 and classify them as removable, poles, or essential singularities. Then find the residue of h ( z ) at each pole....
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Fa11_withsolutionsPM - MASTERS EXAMINATION IN MATHEMATICS...

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