Suppose $f: [0,1] \to [0,1]$ has a periodic orbit ${x_1 < x_2 < x_3 < x_4}$ such that $f(x_i)= x_{i+1}$ for $i < 4$ and $f(x_4) = x_1$. Show that $f$ has periodic points of all periods.
This problem is in a chapter about Sharkovsky Theorem, thus I think I have to use it. I somehow need to show that $f$ has a point of period 3, and then we are done by Sharkovsky. However, how do we show it has a point of period 3?
