Introductory abstract algebra question (computing permutations)

1. Sep 15, 2008

playa007

1. The problem statement, all variables and given/known data
Let A, B be permutations and A = (1 3 5 10)(3 15 8)(4 14 11 7 12 9) and B = (1 14)(2 9 15 13 4)(3 10)(5 12 7)(8 11)
Find AB.

2. Relevant equations

3. The attempt at a solution
I am struggling with finding the product of this permutations and can't quite get the algorithm or process of how to do these things. I've been reading at Dummit and Foote 1.3 on how they find the compute permutation products but I'm having difficulty understanding their algorithm. I think if I put all the subsets of the cycles that are disjoint all on one side (but I'm not clear on how to do this), it might make the problem easier. If one can help with this problem and list all the steps, this would be very very much appreciated.

2. Sep 15, 2008

HallsofIvy

How about just doing what it says? Do you understand what that notation means? That is what is important, not any algorithm!

B = (1 14)(2 9 15 13 4)(3 10)(5 12 7)(8 11)
means that B "changes" 1 into 14 and 14 into 1. B "changes" 2 into 9, 9 into 15, 15 into 13, 13 into 4, and 4 into 2. B "changes" 3 into 10 and 10 into 3. B "changes" 5 into 12, 12 into 7, and 7 into 5. Finally, B "changes" 8 into 11 and 11 into 8.

A = (1 3 5 10)(3 15 8)(4 14 11 7 12 9) should be interpreted in the same way.

Now, put them together. B, as I said before, changes 1 into 14. What does A do to 14? Looking at A I see "(4 14 11 7 12 9)" which tells me that A changes 14 into 11 and so AB changes 1 into 11. Okay, we will start with (1 11 ...). Now what happens to 11? B changes 11 into 8 and, from (3 15 8), A changes 8 into 3. So AB changes 11 into 3 and we have (1 11 3 ...). What about 3? B changes 3 into 10 and A changes 10 into 1:
AB changes 3 into 1 so AB contains (1 11 3). Since 2 is not in that sequence, note that B changes 2 into 9 and, from (4 14 11 7 12 9), A changes 9 into 4: AB changes 2 into 4 so, so far, A= (1 11 3)(2 4 ...

Can you continue from there?