SOLVED 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:
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 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.