if $f$ is of prime order, then can the orbit of $s$ under $f$ have one element?

Question

if $f\in A(S)$ has order $p$, $p$ a prime, show that for every $s\in S$ the orbit of $s$ under $f$ has one or $p$ elements.*

Since the cyclic group generated by $f$ has $p$ elements, therefore $p$ elements for $s$'s orbit. But can you show me when $s$'s orbit only contains one element please?

Answer 1

You can easily prove this by looking at the action of $\;H:=\langle f\rangle\;$, instead of all $\;A(S)\;$, on $\;S\;$ , and then:

$$\forall\,s\in S\;,\;\;\;|\mathcal Orb_H(s)|=[H:H_s]$$

But what are the possible indexes of subgroups of $\;H\;$ ...?

Answer 2

Let $S=\{1,2,3,4,5\}$.

Let $f\in A_5$ be the product of transpositions $f=(1,2)(3,4)$.

Let $s=5$.

The orbit of $s$ under $f$ contains $1$ element.

Hope this helps!