# Thread: Proof regarding cycles of even or odd length

1. ## Proof regarding cycles of even or odd length

I'm having difficulty expressing the proof, even though the statement to be proved seems obvious to me.

Prove: $\alpha^2$ is a cycle iff the length of the cycle $\alpha$ is odd.

What I've got in my head is that if $\alpha=(a,b)$ then $\alpha^2=(a)(b)$. Then if we add another 2 elements c and d, $\alpha=(a,b,c,d)$, making $\alpha^2=(a,c)(b,d)$. The pattern will obviously continue, giving 2 disjoint cycles where the elements alternate for a beginning cycle of any even length. I just need some help assembling my rambling intuitive understanding of the problem into a proper proof.

EDIT: fixed tex tags

2. ## Re: Proof regarding cycles of even or odd length

what you want to prove, is that if α is a n-cycle, with α = (a1 a2 ... an),

then αk sends aj to aj+k (mod n) (you can do this by induction on k).

now if n is even, say n = 2m, then α2 sends:

a1→a3→...→an-1→a1 (since n-1 = 2m -1 is odd)

a2→a4→...→an→a2

that is α2 = (a1 a3 ... an-1)(a2 a4 ... an),

so α splits into 2 disjoint m-cycles.

but if n is odd, say n = 2m+1, so that n-2 is odd, then:

α2 = (a1 a3 .... an-2 an a2 a4 ... an-1),

which is an n-cycle.