# Find the order of a k cycle in group Sn

## Homework Statement

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

## Answers and Replies

Looks fine to me. I guess you're looking for a more concise way of stating this fact though? You could show that for any $i \in \{1,\ldots, k \}$ that
$$f^n(a_i) = a_{i + n \mod k}$$
which is just what you said in more words in your post. This holds for any i, so $f^k(a_i) = a_{i+k \mod k} = a_{i}$, and we conclude that the order of f divides k. All you have to claim is that k is in fact the smallest number and hence is the order. But I think that this is evident.