Periodic Points of a Continuous $f:S^1\to S^1$ Question: Let $f : S^1 \to S^1$ be continuous. Suppose $f$ has a fixed point and a periodic point of prime period $3$. Then does it have to have a periodic point of prime period $2$?
Motivation: I am currently taking a dynamical systems course and while I was reviewing my lecture notes I came across this question. I had thought about this before, but could not come up with an answer. But I remember the professor saying that the answer is "no" (though my memory might be inaccurate). Any ideas or hints are appreciated.
 A: Let's say that your goal is to prove that the answer is indeed "no" (with the pleasant side effect of giving evidence that your memory is indeed accurate).
What you need to do is to take a four point subset $\{w,x_1,x_2,x_3\} \subset S^1$ and to construct a function $f : S^1 \to S^1$ such that 
$$f :
\begin{cases}
w &\mapsto w \\
x_1 &\mapsto x_2 \\
x_2 &\mapsto x_3 \\
x_3 &\mapsto x_1
\end{cases}
$$
and such that $f$ has no point of prime period $2$.
The function $f$ has not yet been completely described: you must still say how $f$ is extended to each of the four complementary subsegments of $\{w,x_1,x_2,x_3\}$. Your goal is to define the extensions of $f$ on those four subsegments so that $f$ has no point of prime period $2$. And even if this goal turns out to be impossible because the answer is actually "yes", you may happen upon disproof by the mere act of looking for proof.
As hinted by @mathstudent, Sharkovskii's Theorem is a helpful guide here, because it tells you that if you are so careless as to extend $f$ so that $f^{-1}(w) = \{w\}$ then $f$ will, indeed, have a point of prime period $2$ (I am implicitly assuming that you will not be so careless as to extend $f$ discontinuously).
So, start working through the cases, keeping these questions in mind: What are the choices for defining $f$ on each of the four complementary subsegments? How does the requirement that $f^{-1}(w) \ne \{w\}$ restrict those choices? And with that restriction, is it possible to make the choices so that $f$ has no point of prime period $2$? Or, on the other hand, will you be forced to create a point of prime period $2$ no matter how you make the choices?
Can you take it from here?
