- #1
PcumP_Ravenclaw
- 106
- 4
Hello everybody,
https://www.physicsforums.com/showthread.php?t=768109
Please see above post for the questions after scrolling down fully.
My answers to questions
1) 4 7 9 2 6 8 1 5 3 is the unique functtion and (1 4 2 7)(3 9)(5 6 8) are the permutations.
2) I don't understand the underlying principle! is (1 2 3 4) the set or the unique function?? what is (1 2 3 4)?? the permuations are not disjoint because they have a common element 1 in them?? I just randomly searched for a way to do it and I found this...
(1 2 3 4) = (1 4)(1 3)(1 2) we can see this as 3 permuations that are commuting with each other but they are not disjoint so a*b != b*a. 1 goes to 2 and 2 goes to no other so (1 2) then 2 goes to 1 and 1 goes to 3 in the second permutation so now it is (1 2 3). 3 goes to 4 in the third permutation and so now it is (1 2 3 4)
3) (1 8 5 4 9)(2 7 6 3 10) ... why are the asking again to write in 2 cycles?? what is p-1 AND what is the identity I? can you please give an example of both??
4) Again I managed to answer this question through trial and error method! but did not understand the underlying princicple!
I don't know what is I? let's just use the definition from group theory. what every binary operation with I is what ever.. The unique functions for the permuations are give below
I, 1 2 3 4, 1 2 3 4, 1 2 3 4
I, 2 1 4 3, 3 4 1 2, 4 3 2 1
a, b , c , d
each element is named a b c & d...
I tried ALL 3 combinations b*c = d , c*d = b, b*d = c they are all within the set so closed axiom is followed...
The inverse is the the element itself?? I Identity is I?? Please explain the inverse
5) I don't understand question 5!
https://www.physicsforums.com/showthread.php?t=768109
Please see above post for the questions after scrolling down fully.
My answers to questions
1) 4 7 9 2 6 8 1 5 3 is the unique functtion and (1 4 2 7)(3 9)(5 6 8) are the permutations.
2) I don't understand the underlying principle! is (1 2 3 4) the set or the unique function?? what is (1 2 3 4)?? the permuations are not disjoint because they have a common element 1 in them?? I just randomly searched for a way to do it and I found this...
(1 2 3 4) = (1 4)(1 3)(1 2) we can see this as 3 permuations that are commuting with each other but they are not disjoint so a*b != b*a. 1 goes to 2 and 2 goes to no other so (1 2) then 2 goes to 1 and 1 goes to 3 in the second permutation so now it is (1 2 3). 3 goes to 4 in the third permutation and so now it is (1 2 3 4)
3) (1 8 5 4 9)(2 7 6 3 10) ... why are the asking again to write in 2 cycles?? what is p-1 AND what is the identity I? can you please give an example of both??
4) Again I managed to answer this question through trial and error method! but did not understand the underlying princicple!
I don't know what is I? let's just use the definition from group theory. what every binary operation with I is what ever.. The unique functions for the permuations are give below
I, 1 2 3 4, 1 2 3 4, 1 2 3 4
I, 2 1 4 3, 3 4 1 2, 4 3 2 1
a, b , c , d
each element is named a b c & d...
I tried ALL 3 combinations b*c = d , c*d = b, b*d = c they are all within the set so closed axiom is followed...
The inverse is the the element itself?? I Identity is I?? Please explain the inverse
5) I don't understand question 5!
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