Cycles

: 136.330 Intro to Algebra: Cycles
   By Manon on Tuesday, October 01, 2002 - 05:03 pm: Edit Post

I'm trying to figure out how to solve a product of disjoint cycle where some of the cycle are the inverse such as (2 4)(2 3 5)^-1. If you could explain the procedure to me again it would be great! Thanks


   By Sasho on Tuesday, October 01, 2002 - 05:52 pm: Edit Post

I have moved your question; please use "Create New Conversation" button more often. Otherwise the topics become too long.

Anyhow; here is what could (or should) be done in your problem. ( I am assuming that the question is to write (2 4)(2 3 5)-1 as a product of disjoint cycles).

Step 1. (we find (2 3 5)-1) Notice that 2 -> 3 , 3-> 5 and 5-> 2 via the mapping (2 3 5) (while the other numbers are fixed); So, the inverse does the following (follow the arrows in the opposite direction): 3->2, 5->3, 2->5. This is good enough. If we wish we can write it as a cycle: (3 2 5).

Step 2. (Find the product (2 4)(2 3 5)-1). Now that we know from step 1 that (2 3 5)-1=(3 2 5), we can write that product as (2 4)(3 2 5). The numbers we do not see are, of course, fixed.

Step 3. (write it as a product of disjoint cycles) start with 2 (we could start with any other); the first cycle will be (2 ...); to fill in the dots we need to see where 2 is mapped. 2 is sent to 5 via (3 2 5), and 5 in turn is sent to 5 via (2 4); so, the next number in the first cyle is 5; so the first cycle is (2 5 ...) with the ... still not known. Next we find where 5 goes: 5 goes to 3 via (3 2 5), and 3 goes to 3 via (2 4). So, our first cycle is (2 5 3 ...) with the dots still not known. Keep going: 3 is sent to 2 via (3 2 5) and 2 is sent to 4 via (2 4). So, we have (2 5 3 4 ...). In the next step you check that 4 is mapped to 2 via the (given) product (2 4)(3 2 5), and since 2 is the first element in our cycle, we close it. Final answer: (2 5 3 4) (your problem happens to yield only one single cycle).

Hope this helps.


   By Manon on Tuesday, October 01, 2002 - 08:02 pm: Edit Post

This helps a lot
Thanks...


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