MTH633 Assignment 1 Solution and Discussion


  • Cyberian's Gold

    Assignment # 1 MTH633 (Fall 2019)

    Total Marks: 10
    Due Date: December 06, 2019.

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    1. Show that the parity of a permutation is even or odd, but not both.
    2. Write down the elements of group under composition

  • Cyberian's Gold

    @zareen said in MTH633 Assignment 1 Solution and Discussion:

    Write down the elements of group under composition

    Test 0: Is the operation associative? If not, then you’re done. 𝑇(𝑆)

    is not a group. If the operation is associative, then proceed to…

    Test 1: Does the set 𝑇(𝑆)
    contains an identity? If you have a candidate function in 𝑇(𝑆), then what equations must it satisfy? Does it? If not, then 𝑇(𝑆)

    is not a group. If you do have an identity, then proceed to…

    Test 2: Does every element of 𝑇(𝑆)
    have an inverse? Given an arbitrary function in 𝑓∈𝑇(𝑆), can you write down its inverse 𝑓−1∈𝑇(𝑆)? What equation must 𝑓 and 𝑓−1 satisfy? (Hint: you need the identity function.) If any function fails to have an inverse, then 𝑇(𝑆)

    is not a group. If every function does have an inverse, then…

    Congratulations! Your set 𝑇(𝑆)
    is a group.
    Reff


  • Cyberian's Gold

    @zareen said in MTH633 Assignment 1 Solution and Discussion:

    Write down the elements of group under composition

    Test 0: Is the operation associative? If not, then you’re done. 𝑇(𝑆)

    is not a group. If the operation is associative, then proceed to…

    Test 1: Does the set 𝑇(𝑆)
    contains an identity? If you have a candidate function in 𝑇(𝑆), then what equations must it satisfy? Does it? If not, then 𝑇(𝑆)

    is not a group. If you do have an identity, then proceed to…

    Test 2: Does every element of 𝑇(𝑆)
    have an inverse? Given an arbitrary function in 𝑓∈𝑇(𝑆), can you write down its inverse 𝑓−1∈𝑇(𝑆)? What equation must 𝑓 and 𝑓−1 satisfy? (Hint: you need the identity function.) If any function fails to have an inverse, then 𝑇(𝑆)

    is not a group. If every function does have an inverse, then…

    Congratulations! Your set 𝑇(𝑆)
    is a group.
    Reff


  • Cyberian's Gold

    @zareen said in MTH633 Assignment 1 Solution and Discussion:

    Show that the parity of a permutation is even or odd, but not both.

    One way is to define the sign of a permutation 𝜎 using the polynomial Δ=Π(𝑥𝑖−𝑥𝑗) with 1≤𝑖<𝑗≤𝑛

    .

    It is easy to see that 𝜎(Δ)=Π(𝑥𝜎(𝑖)−𝑥𝜎(𝑗))
    satisfies 𝜎(Δ)=±Δ. Now define the sign by 𝑠𝑖𝑔𝑛(𝜎)=Δ𝜎(Δ)

    With a little more work you can show that this function is a homomorphism of groups, and that on transpositions it return -1. Therefore, if 𝜎=𝜏1⋯𝜏𝑘
    is a way to write 𝜎 as a product of transpositions, we have 𝑠𝑖𝑔𝑛(𝜎)=(−1)𝑘 and so for even permutations (permutations whose sign is 1) k must always be even, and for odd permutations (whose sign is -1) it must always be odd.



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