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Please verify a problem on Groups and permutations

  1. Aug 29, 2014 #1
    Alan F beardon, Algebra and Geometry
    chapter 1

    6. For any two sets A and B the symmetric difference AΔB of A and B is the
    set of elements in exactly one of A and B; thus
    AΔB = {x ∈ A ∪ B : x / ∈ A ∩ B} = (A ∪ B)\(A ∩ B)
    . Let be a non-empty set and let G be the set of subsets
    of (note that G includes both the empty set ∅ and ). Show that G with
    the operation Δ is a group with ∅ as the identity element of G. What is
    A−1? Now let = {1, 2, . . . , 7}, A = {1, 2, 3}, B = {3, 4, 5} and
    C = {5, 6, 7}. By considering A−1 (inverse) and B−1 (inverse), solve the two equations for x
    AΔX = B, and AΔXΔB = C.

    My answer:

    condition to get null set is

    A\A = 0
    make (A U B) same as (A n B) by changing B

    so A^-1 (A inverse) must be A itself so that (A u A) \ (A n A) = Null set

    Doubts : Are the two X the same or different and why do they want to consider A-1 and B-1. I have solved without considering them.

    for AΔX = B
    X = {1 2 4 5}

    and for AΔXΔB = C
    X = {1, 2, 4, 6, 7}

    x is calculated backwards by AΔX = Y

    so Y Δ B = C

    Y = {3,4,6,7}
    AΔX = Y
    {1,2,3}ΔX = {3,4,6,7}

    so X= {1,2,4,6,7}
     
    Last edited: Aug 29, 2014
  2. jcsd
  3. Aug 29, 2014 #2

    HallsofIvy

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    G is the set of subsets of what? That has been dropped from "let G be the set of subsets of " and "G includes both the empty set and ". Do you mean that G is "the set of subsets" of some "universal set", U?

     
  4. Aug 29, 2014 #3

    HallsofIvy

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    G is the set of subsets of what? That has been dropped from "let G be the set of subsets of " and "G includes both the empty set and ". Do you mean that G is "the set of subsets" of some "universal set", U?

    Have you shown that this is a group? What did you get for the identity?

    ??? There is no "X" in you definition. Did you mean the "x" (NOT "X") in
    "AΔB = {x ∈ A ∪ B : x / ∈ A ∩ B}"? Yes, "x" represents a generic member of AΔB and each member must satisfy both relations.
    (And please don't use "A^-1" in one place and A-1 in another to mean the same thing!)

    Which is the same as using "X= A^{-1}ΔY[/tex]

    so Y Δ B = C

    Y = {3,4,6,7}
    AΔX = Y
    {1,2,3}ΔX = {3,4,6,7}

    so X= {1,2,4,6,7}[/QUOTE]
     
  5. Aug 30, 2014 #4
    corrected version:

    Let Ω be a non-empty set and let G be the set of subsets
    of Ω (note that G includes both the empty set ∅ and Ω). Show that G with
    the operation Δ is a group with ∅ as the identity element of G. What is
    A^−1? Now let Ω = {1, 2, . . . , 7}, A = {1, 2, 3}, B = {3, 4, 5} and C = {5, 6, 7}. By considering A^−1 and B^−1, solve the two equations AΔX = B, and AΔXΔB = C.

    So is my answer for X = {1, 2, 4, 5} for AΔX = B and X = {1, 2, 4, 6, 7} for AΔXΔB = C. correct?

    can i also ask if the associative/commutative rule be used to write AΔXΔB as XΔBΔA?? I get the same asnwer for X!!
     
  6. Aug 30, 2014 #5

    haruspex

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    The first part of the question asked you to show that it forms a group. Demonstrating associativity would have been part of that. It did not ask you to show it was commutative. If you want to use commutativity then you need to demonstrate that too.
     
  7. Sep 7, 2014 #6
    Please educate me solutions to these problems

    Hello everybody,

    Please see the attachment for the questions.

    My answers to questions

    1) 4 7 9 2 6 8 1 5 3 is the unique functtion and (1 4 2 7)(3 9)(5 6 8) are the permutations.

    2) I dont understand the underlying principle!! is (1 2 3 4) the set or the unique function?? what is (1 2 3 4)?? the permuations are not disjoint because they have a common element 1 in them?? I just randomly searched for a way to do it and I found this...

    (1 2 3 4) = (1 4)(1 3)(1 2) we can see this as 3 permuations that are commuting with each other but they are not disjoint so a*b != b*a. 1 goes to 2 and 2 goes to no other so (1 2) then 2 goes to 1 and 1 goes to 3 in the second permutation so now it is (1 2 3). 3 goes to 4 in the third permutation and so now it is (1 2 3 4)

    3) (1 8 5 4 9)(2 7 6 3 10) ... why are the asking again to write in 2 cycles?? what is p-1 AND what is the identity I? can you please give an example of both??

    4) Again I managed to answer this question through trial and error method!! but did not understand the underlying princicple!!

    I dont know what is I? lets just use the definition from group theory. what every binary operation with I is what ever.. The unique functions for the permuations are give below

    I, 1 2 3 4, 1 2 3 4, 1 2 3 4
    I, 2 1 4 3, 3 4 1 2, 4 3 2 1
    a, b , c , d

    each element is named a b c & d....

    I tried ALL 3 combinations b*c = d , c*d = b, b*d = c they are all within the set so closed axiom is followed...

    The inverse is the the element itself?? I Identity is I?? Please explain the inverse

    5) I dont understand question 5!!
     

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