Homework Solution 5 PDF

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MATH 3005 Homework Solution Han-Bom MoonHomework 5 SolutionChapter 5. Let α= [1 2 3 4 5 62 1 3 5 4 6]andβ=[1 2 3 4 5 66 1 2 4 3 5].Compute each of the following.(a)α− 1 α− 1 =[1 2 3 4 5 62 1 3 5 4 6](b) βαβα=[1 2 3 4 5 61 6 2 3 4 5](c) αβ αβ=[1 2 3 4 5 66 2 1 5 3 4] Let α=[1 2 3 4 5 6 7 82 3 4 5 1 7...

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MATH 3005 Homework Solution

Han-Bom Moon

Homework 5 Solution Chapter 5.

1. Let α=

"

#

and β =

=

"

1 2 3 4 5 6 2 1 3 5 4 6

#

βα =

"

1 2 3 4 5 6 1 6 2 3 4 5

#

αβ =

"

1 2 3 4 5 6 6 2 1 5 3 4

#

1 2 3 4 5 6 2 1 3 5 4 6

"

1 2 3 4 5 6 6 1 2 4 3 5

#

.

Compute each of the following. (a) α−1 α

−1

(b) βα

(c) αβ

2. Let α=

"

1 2 3 4 5 6 7 8 2 3 4 5 1 7 8 6

#

and β =

"

1 2 3 4 5 6 7 8 1 3 8 7 6 5 2 4

Write α, β, and αβ as (a) products of disjoint cycles; α = (12345)(678), β = (23847)(56), αβ = (12485736) (b) products of 2-cycles. α = (15)(14)(13)(12)(68)(67), β = (27)(24)(28)(23)(56), αβ = (16)(13)(17)(15)(18)(14)(12) 3. Write each of the following permutations as a product of disjoint cycles. (a) (1235)(413) (15)(234) (b) (13256)(23)(46512) (124)(35) or (124)(35)(6) 1

#

.

MATH 3005 Homework Solution

Han-Bom Moon

(c) (12)(13)(23)(142) (1423) 6. What is the order of each of the following permutations? " # 1 2 3 4 5 6 (a) 2 1 5 4 6 3 Let α be the permutation. Then α = (12)(356). α2 = ((12)(356))2 = (12)2 (356)2 = (365), α3 = ((12)(356))3 = (12)3 (356)3 = (12), α4 = (356), α5 = (12)(365), α6 = e. Thus |α| = 6. Indeed, you may find a pattern. If α = α1 α2 · · · αk where αi are disjoint cycles of length ℓi , then the order of α is lcm(ℓ1 , ℓ2 , · · · , ℓk ). " # 1 2 3 4 5 6 7 (b) 7 6 1 2 3 4 5 Let β be the permutation. Then β = (1753)(264). By the same idea, you may find |β| = lcm(4, 3) = 12. 11. Determine whether the following permutations are even or odd. (a) (135) (135) = (15)(13) So it is an even permutation. (b) (1356) (1356) = (16)(15)(13) It is an odd permutation. (c) (13567) (13567) = (17)(16)(15)(13) It is an even permutation. (d) (12)(134)(152) (12)(134)(152) = (12)(14)(13)(12)(15) It is an odd permutation. (e) (1243)(3521) (1243)(3521) = (13)(14)(12)(31)(32)(35) It is an even permutation.

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MATH 3005 Homework Solution

Han-Bom Moon

15. Let n be a positive integer. If n is odd, is an n-cycle an odd or an even permutation? If n is even, is an n-cycle an odd or an even permutation? Note that an n-cycle α = (a1 a2 · · · an ) is a product (a1 an )(a1 an−1 ) · · · (a1 a2 ). So it is a product of (n − 1) 2-cycles. Therefore if n is odd, then it is an even permutation and if n is even, then it is an odd permutation. 34. What cycle is (a1 a2 · · · an )−1 ? The cycle α := (a1 a2 · · · an ) is a function where α(a1 ) = a2 , α(a2 ) = a3 , · · · , α(an−1 ) = an , and α(an ) = a1 . So α−1 (an ) = an−1 , · · · , α−1 (a2 ) = a1 , and α−1 (a1 ) = an . Therefore α−1 = (an an−1 · · · a2 a1 ). 35. Let G be a group of permutations on a set X. Let a ∈ X and define stab(a) = {α ∈ G | α(a) = a}. We call stab(a) the stabilizer of a in G (since it consists of all members of G that leave a fixed). Prove that stab(a) is a subgroup of G. First of all, stab(a) 6= ∅ because e ∈ stab(a). Let α, β ∈ stab(a). Then α(a) = a and β(a) = a. αβ(a) = α(β(a)) = α(a) = a so αβ ∈ stab(a). Also for α ∈ stab(a), α−1 (a) = a because a = α−1 (α(a)) = α−1 (a). Therefore α−1 ∈ stab(a) as well. So stab(a) ≤ G. 57. Viewing the members of D4 as a group of permutations of a square labeled 1, 2, 3, 4 as described in Example 3, which geometric symmetries correspond to even permutations? If you denote all elements of D4 as a permutation on the vertex set {1, 2, 3, 4} as in Example 3, we have R0 = e, R90 = (1234) = (14)(13)(12), R180 = (13)(24), R270 = (1432) = (12)(13)(14), H = (12)(34), V = (14)(23), D = (24), D′ = (13). Therefore R0 , R180, H, and V are even permutations. 69. Prove that every element of Sn (n > 1) can be written as a product of elements of the form (1k). Because all permutations α ∈ Sn is a product of 2-cycles, if each 2-cycle (ij) is a product of 2-cycles of the form (1k), we are able to obtain the conclusion. But (ij) = (1i)(1j )(1i), so we have the result.

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