Fifth Homework Solutions

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$Stanca Mihaela Ciupe - Math Department - Virginia Tech$

$S13 MATH 1205 – Test 1 20 Feb 2013 NAME: CRN: 2 2 4 6 8 10 x 2 ...$

$Boris Kramer - Math Department - Virginia Tech$

$here - Math Department - Virginia Tech$

$Math 2534 Solution to Homework 8 on PMI$

$Math 2534 Solution for Homework on PMI$

so θ is a homomorphism. Suppose g ∈ G, then g = e2kπi/pn for some k,n ∈ Z. Set h = e2kπi/pn+1. Then h ∈ G and θh = g. Therefore θ is surjective. Finally e2πi/p ∈ kerθ, so kerθ �= 1. The result now follows from the first isomorphism theorem. 3. Let G be a group, and suppose there is a homomorphism of G onto S3 (the symmetric group of degree 3) with kernel K. Determine the number of subgroups of G which contain K, and show that exactly three of these subgroups are normal. By the subgroup correspondence theorem, the subgroups containing K are in a oneto-one correspondence with the subgroups of G/K, and this correspondence preserves normality. This means we have to determine the subgroups of S3. Any subgroup of S3 has order dividing 6, so any subgroup not equal to 1 or S3 has prime order and is therefore cyclic. It follows easily that the subgroups of S3 are 1, S3, 〈(12)〉, 〈(23)〉, 〈(31)〉, 〈(123)〉, so there are six subgroups of G which contain K. Finally the normal subgroups of S3 are 1, S3 and 〈(123)〉. We conclude that the number of normal subgroups containing K is 3, as required. 4. 3.5.2 on page 111. Prove that σ 2 is an even permutation for every permutation σ. We may write σ as a product of n transpositions for some positive integer n. Then we may write σ 2 as a product of 2n transpositions, which shows that σ is an even permutation. 5. 4.1.5(a) on page 116. Let S3 act on the set A of 27 ordered triples {(i, j,k) | 1 ≤ i, j,k ≤ 3} by σ(i, j,k) = (σi,σ j,σk). Find the orbits of S3 on A. For each σ ∈ S3, find the cycle decomposition of σ under this action. For each orbit O of S3 acting on these 27 points, pick some a ∈ O and find the stabilizer of a in S3. The size of an orbit divides the order of the acting group, which is here S3, so we are looking for orbits of size 1,2,3 or 6. By considering just the first coordinate, it is easy to see that all the orbits must have size at least 3, so each orbit has size 3 or 6. We will consider S3 as {1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}. Then the orbits are {(1,1,1), (2,2,2), (3,3,3)} {(1,1,2), (2,2,1), (3,3,2), (1,1,3), (2,2,3), (3,3,1)} {(1,2,1), (2,1,2), (3,2,3), (1,3,1), (2,3,2), (3,1,3)} {(1,2,3), (2,1,3), (3,2,1), (1,3,2), (2,3,1), (3,1,2)} {(1,2,2), (2,1,1), (3,2,2), (1,3,3), (2,3,3), (3,1,1)} The stabilizer of (1,1,1) in S6 is {1, (2 3)}, while the stabilizer of all elements in any of the other four orbits is 1. We will let (a,b,c) correspond to 9(a − 1) + 3(b − 1) + c. This means the elements of S are
(1,1,1)=1, (1,1,2)=2, (1,1,3)=3, (1,2,1)=4, (1,2,2)=5, (1,2,3)=6, (1,3,1)=7, (1,3,2)=8, (1,3,3)=9, (2,1,1)=10, (2,1,2)=11, (2,1,3)=12, (2,2,1)=13, (2,2,2)=14, (2,2,3)=15, (2,3,1)=16, (2,3,2)=17, (2,3,3)=18, (3,1,1)=19, (3,1,2)=20, (3,1,3)=21, (3,2,1)=22, (3,2,2)=23, (3,2,3)=24, (3,3,1)=25, (3,3,2)=26, (3,3,3)=27 Element Cycle Decomposition 1 (1) (1 2) (1 14)(2 13)(3 15)(4 11)(5 10)(6 12)(7 17)(8 16)(9 18)(19 23)(20 22)(21 24)(25 26) (1 3) (1 27)(2 26)(3 25)(4 24)(5 23)(6 22)(7 21)(8 20)(9 19)(10 18)(11 17)(12 16)(13 15) (2 3) (2 3)(4 7)(5 9)(6 8)(10 19)(11 21)(12 20)(13 25)(14 27)(15 26)(16 22)(17 24)(18 23) (1 2 3) (1 14 27)(2 15 25)(3 13 26)(4 17 21)(5 18 19)(6 16 20)(7 11 24)(8 12 22)(9 10 23) (1 3 2) (1 20 16)(2 25 15)(3 26 13)(4 21 17)(5 19 18)(6 20 16)(7 24 11)(8 22 12)(9 23 10)
Page 1: Math 4124 Wednesday, March 16 Fifth

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