SOLVED MTH633 Assignment 1 Solution and Discussion
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Assignment # 1 MTH633 (Fall 2019)
Total Marks: 10
Due Date: December 06, 2019.DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
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• 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
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@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 -
@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 -
@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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