I was wondering if someone could help take me through how you'd solve the following permutations:
Sorry if it looks weird, don't know how to create bigger brackets, LHS and RHS are both meant to be within the same sets of brackets not separate.
1. Simplify and find the sign of
(1 2 3 4 5 6) * (1 2 3 4 5 6)
(2 4 3 6 5 1) (1 3 6 2 4 5)
Now I'd know how to solve these individually, the LHS would go to (1246)(3)(5) which would go to (1246)
Similarly I know how to do the RHS but don't know how to begin when multiplying two.
2. Simplify and find the sign of
(2,6)(1,2,3,4,5)(1,4,6)
I'm not sure how to go about simplifying this
I believe the OP is referring to permutation groups/permutation cycles as in this article and this article. There might be more responses if this were in the Linear and Abstract Algebra subforum since this falls under group theory. I'd have to brush up quite a bit to be able to answer the OP's question so I left it for others to answer.
Unfortunately, there is no general rule as to whether a product of permutations is to go "right first then left" or "left first then right". There are text books that use each and I don't know which convention your textbook uses. I am going to treat it like a "composition" of functions and do "right first then left". The right permuation takes 1 to 1 and the the left takes 1 to 2- together, they take 1 to 2. The right permutation take 2 to 3 and the the left takes 3 to 3- together they take 2 to 3. The right permutation take 3 to 6 and then the left takes 6 to 1- together they take 3 to 1. The right permutation takes 4 to 2 and then the right take 2 to 4- together they take 4 to 4. The right permutation take 5 to 4 and then the right takes 4 to 6- together they take 5 to 6. The right permutation takes 6 to 5 and then the left takes 5 to 5- together they take 6 to 5.
That is, the two permutations (going right to left) give 1->2, 2->3, 3->1, 4->4, 5->6, 6->5 which can be written
(1 2 3 4 5 6)
(2 3 1 4 6 5)
That can also be written in "cycle" notation as (123)(56).
It can be done with individual transpositions as
(1 2 3 4 5 6)
(2 1 3 4 5 6)
(2 3 1 4 5 6)
(2 3 1 4 6 5)
requiring 3 transpostions. That is an odd permutation so its sign is -1.
By "solving" you mean writing in "cycle" notation?Now I'd know how to solve these individually, the LHS would go to (1246)(3)(5) which would go to (1246)
Similarly I know how to do the RHS but don't know how to begin when multiplying two.
Yes, the left permutation is (1246) and the right is (23654). Again, using the "right first" convention (your class may be using "left first") I see that there is no "1" in (23654) so I go immediately to (1246) and see that 1->2. The first thing I write is "(12 ". Now, on the right I see "23" and on the left there is no "3" so 2-> 3. Now, I have "(123 ". On the right I see "36" and on the left 6 is the end of the list so 6->1. 3->1, the beginning of the cycle, so that cycle is complete- "(123)".
The next number is 4 and on the right 4 goes back to the beginning of the cycle, 2. On the right I see "24". That is, 4->4 and we don't need to write that. The next number is 5 and on the right I see "54" while on the left is "46". Together those give 5-> 6 or "(56 ". Of course then I see, on the right, "65" while there is no "5" on the left. That tells me that 6->5 as had to happens since there were no other numbers. Together we have (123)(56). Because those are disjoint cyles, the order in not important. Note that since (123) contains 3 numbers there are 3-1= 2 transpositions. Since (56) contains only two numbers it is a transposition itself: 2- 1= 1 transpositions. Together there are 2+ 1= 3 transpostions so this is an odd permutation as I said before.
Again working from the right, 1> 4 and then 4->5 so start with "(1 5 ". On the right there is no 5 but in the middle cycle, 5-> 1 and there is no 1 on hte right so (1 5) is a cycle. The next number is 2. There is no 2 in the right cycle so I look in the middle and see "2,3". There is no 3 in the last cycle so we have "(2 3 ". 3->4 in the middle cycle and there is no 4 in the last so "(2 3 4 ". In the first cycle, we have "4,6". There is no 6 in the middle cycle but the first is (2,6) so we are back to 2. The complete cycle is (2 3 4 6). That completes all the numbers. Simplified to disjoint cycles, this is (1 5)(2 3 4 6). The first cycle has 2 numbers so is 1 transposition and the second has 4 numbers so has 4- 1= 3 transpositions. Together they have 3+1= 4 transpositions so this is an even transposition and has sign "+".2. Simplify and find the sign of
(2,6)(1,2,3,4,5)(1,4,6)
I'm not sure how to go about simplifying this
Again, check to see which convention your class is using.