# Can someone please check this proof involving permutations and cycles?

1. May 7, 2012

### Syrus

1. The problem statement, all variables and given/known data

Prove: If σ is a cycle of odd length, then σ2 is a cycle.

2. Relevant equations

N/A

3. The attempt at a solution

Proof: Assume σ is a cycle of odd length. Then let us model σ as (1 2 3 4 ... 2k +1) for some integer k [assuming here that the fixed elements of σ are those such that 2k + 1 < i]. Note then that σ2 = (1 2 3 4 ... 2k +1)(1 2 3 4 ... 2k +1), which we may otherwise define as σ2(i) = {i + 2, i ≤ 2k - 1; 1, i = 2k; 2, i = 2k + 1; i, 2k + 1 < i}. Hence, the elements 1,2,3,4,...,2k + 1 of σ remain fixed in a cycle, and so σ2 is indeed also a cycle.

2. May 7, 2012

### tiny-tim

Hi Syrus!
How does that prove it's a cycle (and not eg 2 cycles)?

Doesn't your proof also apply to even lengths?

3. May 8, 2012

### Syrus

I think i understand what you're implying. I have now taken into account the fact that σ is odd- does this seem acceptalbe?

Proof: Assume σ is a cycle of odd length. Then let us model σ as (1 2 3 4 ... 2k + 1) for some integer k [assuming here that the fixed elements of σ are those n such that 2k + 1 < n]. Note then that σ2 = (1 2 3 4 ... 2k +1)(1 2 3 4 ... 2k + 1). Then σ2 can be alternatively represented by (1 3 5 7 ... 2k + 1 2 4 6 8 ... 2k). Since σ(2k) = 2k + 1, σ(2k + 1) = 1, and σ(1) = 2, it follows that σ2(2k) =1 and σ2(2k + 1) = 2. Hence, the elements 1,2,3,4,...,2k + 1 of σ remain fixed in a cycle, and so σ2 is indeed also a cycle.

4. May 8, 2012

### tiny-tim

Hi Syrus!

You could have stopped here
… you've actually specified the cycle whose existence you had to prove (with all the elements distinct).