Let π=(1 2 5 7)(2 3 4)(1 3 5)(2 3 7 8 9)
I need to write π in the standard form where 1,2,...,9 is on top and then the image of these numbers is below them.
I realize that these cycles are non-disjoint, and that's what's throwing me off. It seems like 1 connects with 2 and 1 connects with 3. But I know that 1 can only have one image. So I'm kind of lost here. Help please!
Ok, I understand that method. But I thought that that is what you want to do when you want to write π as a product of disjoint cycles. Writing π as a product of disjoint cycles would be (1 4 5 2 7 8 9 3).
SO, back to my original question, if I want to write π like π=1 2 ....9 with the images underneath, I'm guessing the images come from my product of disjoint cycles above? In other words, would π written in standard form be...
π=1 2 3 4 5 7 8 9
...4 7 1 5 2 8 9 3
Also, I know that the order for a disjoint permutation is ord(π1, π2...)=lcm(ord(π1), ord(π2),...), but I don't know how to find the order for a non-disjoint permutation. I would appreciate an explanation of how to find it.
the procedure goes like this: in the product, what is the image of 1? well 1 isn't moved in the cycle (2 3 7 8 9), so we move to the next one in the compositional order, (1 3 5). in this cycle, 1 is moved, and has image 3. so in the NEXT cycle, we check to see if 3 is moved or not. and it is, 3 is moved to 4 in (2 3 4). since 4 is not moved by (1 2 5 7), we are done calculating the image of 1. whew! only 8 more images to check. now, assuming your disjoint cycle is correct (i haven't checked, but it looks like it may be), then the "standard" way of writing it as a 1-1 function by listing the image set under the domain set is correct.
in general, it is easier to write a permutation as a product of disjoint cycles to find its order. given the standard form, you just write (1 f(1) f(f(1))....f^-1(1)) and then start the second cycle with the first element not moved by the first cycle, until you've exhausted the numbers involved.