Prove that a k-cycle in the group Sn has order k.
The Attempt at a Solution
I'm mostly confused on how to write this in math notation. I know it will have order k because a1 → a2 → a3 .... ak-1 → ak → a1 if we do the compositions K times. and so everything will get mapped back to itself and hence we have the identity permutation.
Let f = (a1,a2,a3...,ak)
then f(a1) = a2 this is the original permutation
then if we do the compoisition of f o f we have
a1 → a3 , a2→ a4. .. etc. so f^2(a1) = a3
then if we do f o f o f we have a1 → a4 so
f^3(a1) = f(a3) = a4
f^k(a1) = f(ak) = a1
and this process would hold for any element in f, so we see that f^k =(1) , and so f has order of K. therefore any k cycle group in Sn has order K.
Any help is appreciated.Thanks