MTH633 Assignment 1 Solution and Discussion

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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 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 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.