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MTH633 Assignment 1 Solution and Discussion

Assignment # 1 MTH633 (Fall 2019)
Total Marks: 10
Due Date: December 06, 2019.DONβT MISS THESE: Important instructions before attempting the solution of this assignment:
β’ To solve this assignment, you should have good command over 1116 lectures or module 4059.
β’ Upload assignments properly through LMS, No Assignment will be accepted through email.
β’ Write your ID on the top of your solution file.
β’ Donβt use colorful back grounds in your solution files.
β’ Use Math Type or Equation Editor etc for mathematical symbols if needed.
β’ You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
β’ Make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
β’ Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments. Show that the parity of a permutation is even or odd, but not both.
 Write down the elements of group under composition

@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 youβre done. π(π)
is not a group. If the operation is associative, then proceed toβ¦
Test 1: Does the set π(π)
contains an identity? If you have a candidate function in π(π), then what equations must it satisfy? Does it? If not, then π(π)is not a group. If you do have an identity, then proceed toβ¦
Test 2: Does every element of π(π)
have an inverse? Given an arbitrary function in πβπ(π), can you write down its inverse πβ1βπ(π)? What equation must π and πβ1 satisfy? (Hint: you need the identity function.) If any function fails to have an inverse, then π(π)is not a group. If every function does have an inverse, thenβ¦
Congratulations! Your set π(π)
is a group.
Reff

@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 youβre done. π(π)
is not a group. If the operation is associative, then proceed toβ¦
Test 1: Does the set π(π)
contains an identity? If you have a candidate function in π(π), then what equations must it satisfy? Does it? If not, then π(π)is not a group. If you do have an identity, then proceed toβ¦
Test 2: Does every element of π(π)
have an inverse? Given an arbitrary function in πβπ(π), can you write down its inverse πβ1βπ(π)? What equation must π and πβ1 satisfy? (Hint: you need the identity function.) If any function fails to have an inverse, then π(π)is not a group. If every function does have an inverse, thenβ¦
Congratulations! Your set π(π)
is a group.
Reff

@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 π using the polynomial Ξ=Ξ (π₯πβπ₯π) with 1β€π<πβ€π
.
It is easy to see that π(Ξ)=Ξ (π₯π(π)βπ₯π(π))
satisfies π(Ξ)=Β±Ξ. Now define the sign by π πππ(π)=Ξπ(Ξ)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 π=π1β―ππ
is a way to write π as a product of transpositions, we have π πππ(π)=(β1)π 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.