Finite groups of functions under function composition

Question

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.

Answer 1

Depending on exactly what your requirements are, there does not exist such a function. First let's suppose you're working with rational functions $f(z) = \frac{p(z)}{q(z)}$ where $z$ is a complex parameter and $p, q$ have no common roots. In order for $f$ to be invertible by another rational function, it needs to be a bijective function from the Riemann sphere to itself. Now:

• If $\deg p > 1$ then $f^{-1}(0)$ consists of more than one point, and
• If $\deg q > 1$ then $f^{-1}(\infty)$ consists of more than one point.

It follows that in order for $f$ to be invertible we must have $f(z) = \frac{az + b}{cz + d}$; moreover we need $ad - bc \neq 0$, and this is sufficient.

The resulting group is very well-understood. It is the group of Möbius transformations or, equivalently, the projective special linear group $\text{PSL}_2(\mathbb{C})$, and all of its finite subgroups are known. Remarkably, they are described by Dynkin diagrams (see e.g. this MO question). Explicitly they are:

• the finite cyclic groups,
• the finite dihedral groups,
• the tetrahedral group,
• the octahedral group, and
• the icosahedral group.

The good news is that there are elements of order $5$. The bad news is that they cannot be realized over $\mathbb{Q}$.

To see this, note that $\text{PSL}_2(\mathbb{C})$ admits a 2-to-1 homomorphism $\text{SL}_2(\mathbb{C}) \to \text{PSL}_2(\mathbb{C})$; thus if $f(z) = \frac{az + b}{cz + d}$ (written so that $ad - bc = 1$) is an element of order $5$ or $7$, its preimages in $\text{SL}_2(\mathbb{C})$, one of which is given by the matrix $$M = \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right],$$

must have order $5, 7, 10$ or $14$. Since it has determinant $1$, its eigenvalues must be a primitive root of unity $\zeta_n$ (where $n = 5, 7, 10, 14$) and $\zeta_n^{-1}$. This means that its trace must be $$\zeta_n + \zeta_n^{-1} = 2 \cos \frac{2 \pi i}{n}$$

but it must also be $a + d$, which is rational. Since $\zeta_n + \zeta_n^{-1}$ is a sum of algebraic integers, it is an algebraic integer, hence is rational if and only if it is an integer. But $2 \cos \frac{2 \pi i}{n}$ is an integer if and only if it's equal to $2, 1, 0, -1, -2$, and this gives $n = 1, 2, 3, 4, 6$. Thus we cannot have $n = 5, 7, 10, 14$ or any number greater than $6$ for that matter.

The above argument is closely related to the crystallographic restriction theorem.

Answer 2

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.