What is the order of $f(x) = \frac{2}{2-x}$ as a permutation?

Question

$f(x) = \frac{2}{2-x}$ defines a permutation of the set $A={\bf R}\setminus \{0,1,2\}$. What is its order in the permutation group $S_A$?

I don't know how to find orders of functions. Please help!

Answer 1

I don't think there is any way to do it much smarter except calculating the compositions.

Unless I've miscalculated myself, the order is $4$. You can calculate that a bit faster by first calculating $g=f^2$ (this is composition, not multiplication) and then $g^2=f^4$.

Answer 2

For the purpose at hand, tomasz's answer resolves the OP's question, but I thought I'd just add a note that there is a "smart" way to compute the order of a function if it's a linear fractional transformation of the form $f(x)=(ax+b)/(cx+d)$ with real coefficients: compute the eigenvalues of the $2\times2$ matrix

$$\pmatrix{a&b\\c&d}$$

If they are of the form $\lambda=re^{\pm\pi im/n}$ with $r$ real and the fraction $m/n$ in reduced form (i.e., $\gcd(m,n)=1$), then the function has order $n$. (That is, $f(f(\cdots f(x)))=x$, where there are $n$ $f$'s on the left hand side.)

In the OP's case, the matrix is

$$\pmatrix{0&2\\-1&2}$$

and the eigenvalues are $\sqrt2e^{\pm\pi i/4}$, so the order is $4$.