Prove that the tent map has exactly nine 6-cycles. Prove that the tent map
$$T(x)= \begin{cases}2x& \text{if} &0 \le x \le 1/2\\2-2x& \text{if}&1/2<x \leq 1\end{cases}$$
has exactly nine 6-cycles.
This is my first proofs class and we have yet to prove anything in class. My professor gave us this problem to do by Monday and I honestly have no idea of where to start or what to do.
 A: There are $2^n$ equations for $T^n(x)$, and so there are $2^n$ fixed points. However, some of the fixed points are in the same cycle. So for $T^3(x)$, the fixed points are $\left\{0,\frac29,\frac27,\frac49,\frac47,\frac69=\frac23,\frac67,\frac89\right\}$, and the cycles for $T^3(x)$ are $\left\{\left\{0\right\},\left\{\frac29,\frac49,\frac89\right\},\left\{\frac27,\frac47,\frac67\right\},\left\{\frac69=\frac23\right\}\right\}$. So there are 4 period-3 orbits for $T(x)$, but only 2 with a prime period of 3, and thus only 2 3-cycles. 
**EDIT**:
Prove that the tent map has exactly nine 6-cycles.
$$T(x)=\left\{\begin{matrix}2x \qquad \ \  \ \ \ 0\le x\le\frac12 \\ 2-2x \quad \frac12< x\le1\end{matrix}\right.$$


*

*$T^n(x)$ has $2^n$ piecewise functions, $n \in N$ by the definition of an iterating function and by $T(x)$. 

*$T^n(x)$ has $2^n$ fixed points where $T^n(x)=x$. By (1). 

*Iff $T^n(x)=x$, then $x$ is an element of a period-$n$ orbit of $T(x)$. By definition of orbit. 

*The fixed points of $T^n(x)$ comprise all elements of the period-$n$ orbits of $T(x)$. By (3). 

*Given $m<n$ where $m,n \in N$, iff $\frac mn \in N$, then $m$ is a factor of $n$. The factors of $n$ are $\{m_1,m_2,...,m_k\}$ where $k \in N$. By the definition of factor. 

*$T^n(x)=T^{m_i}(T^{m_j}(x))$, where $m_i \cdot m_j=n$ and $i,j \in N$. By iterating function.

*The fixed points of $T^{m_k}(x)$ are also fixed points of $T^n(x)$. By (6). 

*Given $s$ unique fixed points of $T^{m_k}(x)$, there are $2^n-s$ fixed points of $T^n(x)$ which are not fixed points of any $T^{m_k}(x)$. By (7) and (2).

*The fixed points of $T^n(x)$ which are not fixed points of $T^{m_k}(x)$ are all the elements of period-$n$ orbits of $T(x)$ with prime period $n$. By (7) and (3). 

*There are $n$ unique elements to each $n$-cycle of $T(x)$. By definition of $n$-cycle.

*There are exactly $\frac {2^n-s}{n}$ $n$-cycles of $T(x)$. By (10). 

*For $T^6(x)$, there are $\frac {2^6-s}{6}$ 6-cycles of $T(x)$. By (11). 

*$s$=number of unique fixed points of $\{T^1(x),T^2(x),T^3(x)\}=2+2+6=10$. By (8) and (5). 

*There are $\frac {2^6-10}{6}$ 6-cycles of $T(x)$. $$\frac {2^6-10}{6}=\frac {64-10}{6}=\frac {54}{6}=9$$ By (13),(11),(12).

*There are exactly nine 6-cycles of $T(x)$. By (14). 
Hope that's more comprehensive and helpful.
A: The base $2$ Lyndon words of length $6$ count the cycles, so you can write out the $64$ binary strings of length two.
Then put aside the ten strings which are repetitions of $1,2,3$-cycles i.e. $000000,111111,010101,101010,100100,010010,001001,110110,011011,101101$.
Then group the rest into strings which are cycles of each other and you will have nine equivalence classes each containing six rotations of the same string.  The nine classes are in bijection with the nine cyclic sets you seek.
