 Solution

Write the period-three points in increasing order and take the two intervals between consecutive points. For either possible cyclic ordering, the interval-covering transition graph is, after interchanging its vertices,The loop at the first vertex gives a fixed point. For every $n\ge 2$, the closed itineraryhas least period $n$. The interval-covering periodic-orbit theorem supplies a point with that itinerary and therefore a cycle of least period $n$. Thus $F$ has periodic orbits of every positive period. This is the period-three case of the Sharkovsky theorem.