# cycles

• Nov 22nd 2008, 07:38 PM
safecracker
cycles
I'm working on a midterm review sheet for my MATH 417 class. It is noted that the midterm (which will be held at 10am Monday) will look very much like the review. I'm having a lot of trouble solving one question:

"Let b be a cycle of length at least 3 in Sn­. Prove that b2 is a cycle if and only if the length of b is odd."

Any ideas?
• Nov 22nd 2008, 07:45 PM
ThePerfectHacker
Quote:

Originally Posted by safecracker
I'm working on a midterm review sheet for my MATH 417 class. It is noted that the midterm (which will be held at 10am Monday) will look very much like the review. I'm having a lot of trouble solving one question:

"Let b be a cycle of length at least 3 in Sn*. Prove that b2 is a cycle if and only if the length of b is odd."

Any ideas?

Your question makes almost as much sense as a magician in front of blind people.
I think you want to say $\displaystyle \beta \in S_n$ is a cycle where $\displaystyle n\geq 3$?
• Nov 22nd 2008, 08:09 PM
safecracker
I just copied and pasted it from the pdf. But yes, what you wrote is what was is meant by it. Also note that in the second sentence Beta is squared.
• Nov 22nd 2008, 08:25 PM
ThePerfectHacker
Quote:

Originally Posted by safecracker
I just copied and pasted it from the pdf. But yes, what you wrote is what was is meant by it. Also note that in the second sentence Beta is squared.

Instead of writing out a full proof I think it is easiet to see what is going on between the parity by writing out a few example. First, we will consider when the length of $\displaystyle \beta$ is even:

If $\displaystyle \beta = (1234)$ then $\displaystyle \beta^2 = (13)(24)$.
If $\displaystyle \beta = (123456)$ then $\displaystyle \beta^2 = (135)(246)$.
If $\displaystyle \beta = (12345678)$ then $\displaystyle \beta^2 = (1357)(2468)$.

Now look what happens when $\displaystyle \beta$ is odd:

If $\displaystyle \beta = (123)$ then $\displaystyle \beta^2 = (132)$.
If $\displaystyle \beta = (12345)$ then $\displaystyle \beta^2 = (13524)$.
If $\displaystyle \beta = (1234567)$ then $\displaystyle \beta^2 = (1357246)$.

It should be clear now how parity plays a role. When we have even number of elements in the cycle then the cycle terminates in the middle and returns to begining and that is why we get two disjoint cycles of halved length. And if we have an odd number of elements then the cylce manages to fully complete itself.
• Nov 22nd 2008, 09:01 PM
safecracker
Thanks, that makes a lot of sense. I think I was confused on the wording of the question.