TITLE: Finite groups of functions under function composition
QUESTION [13 upvotes]: Over the years I have done many questions along the lines of the following:
"Given functions $\phi, \theta$ (usually defined on $\mathbb{R}$ or $\mathbb{C}$, or a suitable subset of $\mathbb{R}$ or $\mathbb{C}$) prove that the collection of all functions obtained from $\theta$ and $\phi$ by function composition form a group G." (Frequently G is $\mathbb{Z}/4 \mathbb{Z}$ or $S_3$.)
A typical example might be functions $\theta:x\mapsto 1-x$ and $\phi:x\mapsto \frac{1}{x}$, (defined on the set of non-zero reals) generating a group isomorphic to $S_3$.
When I have been inventing questions for my students, rather than simply copying examples from previous exam papers, I have sometimes wondered if it is possible to create a similar example with a function $\psi$ which has order 5 in the group - just for a bit of variety! However a crucial restriction is that I need the functions to be simple algebraic functions (something like $x\mapsto \frac{ax+b}{cx+d}$ with $a, b, c, d\in \mathbb{Z}$), which rules out things like rotations of $\mathbb{C}$ through an angle of $2\pi/5$.
My question is then: does anyone know of such a function, or has anyone come across a similar exam question which gives an element of order 5 (or 7 for that matter ...) arising from such a simple type of function?
I have tried investigating the possibilities at various times, and have easily found functions which have order 2, 3, 4, but never one of order 5 or 7.
PS There is no great urgency here, as I have now retired from teaching this sort of stuff.

REPLY [2 votes]: You can look for a rotation in $\mathbb{H^2}$. For example suposse fixed $i$ and then $-i$. The expression looking for $w=w(z)$ with
$$\frac{w+i}{w-i}=k\frac{z+i}{z-i}$$
Then $$w(z)=\frac{(k+1)z+(k-1)i} {(k-1)z+(k+1)i}$$
You want $w^5(z)=z$. In matrix notation you need that
$$\left( \begin{array}{ll}
 k+1 & k-1\\
 k-1 & k+1
\end{array}\right)^5$$ be a scalar matrix. As
$$\left( \begin{array}{ll}
 k+1 & k-1\\
 k-1 & k+1
\end{array}\right)^5=\left( \begin{array}{ll}
 16 k^5+16 & 16 k^5-16\\
 16 k^5-16 & 16 k^5+16
\end{array}\right)$$
Then $k^5=1$, for example $k=e^{2\pi i/5}$. If you want $w^7(z)=z$, in the same way $k^7=1$. 
Certainly, it is not a simply and closed expression with integer numbers.