Given the map $\tilde{f}:[1,13]\to[1,13]$, and $\tilde{f}(x)=\begin{cases} f(x)+8;& 1\le x\le 5\\ x-8;& 9\le x \le 13\\ \end{cases}$, which looks like the figure: 
Prove that the map has a 10-cycle but no 6-cycles. I have been able to show that we have the 10-cycle: $f:\{11,3,12,4,10,2,13,5,9,1\}$, but now I'm stuck on what to do. I considered taking the approach that the cycle is odd and hence would have to have two numbers $\tilde{f}(a)=b$ and $\tilde{f}(b)=a$, but I'm not sure this is valid. Please help.
A 6-cycle would have to be found among the remaining $13-10$ numbers.