= 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,
$$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}.
$$
The loop at the first vertex gives a fixed point. For every $n\geq2$, the closed itinerary
$$
0^{n-1}1
$$
has 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>.
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