Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
IOW: what orders can elements of $\mathrm{SL}_2(\Bbb Z)$ have? – anon Jun 14 '12 at 22:16
Not exactly, as I would allow things like simple quadratics - the crucial thing is that the functions have to be algebraically simple enough so that students could reasonably be expected to work out several compositions under exam conditions. (Sorry to be a bit vague about what "simple" means here!) – Old John Jun 14 '12 at 22:17
Is it possible for a (lowest terms) rational function with a numerator or denominator of degree greater than 1 to have finite order under composition? You prove something like that is impossible when finding the automorphisms of k[x] or k(x) or whatever. – Jack Schmidt Jun 14 '12 at 22:22
... but, isn't any rotation of $\mathbb{C}$ (about the origin) given by an algebraic rule of the form $x \mapsto ax$, for some $|a|=1$? I guess I want a better idea of what "simple" means. Complex multiplication (as opposed to real) isn't simple? Could that be it? – leslie townes Jun 14 '12 at 22:29
FWIW if an $2 \times 2$ integer matrix has finite order, the order must be $1$, $2$, $3$, $4$, or $6$. (The minimal polynomial of such a matrix must divide some polynomial of the form $x^n - 1$, and hence be a product of cyclotomic polynomials. Degree considerations rule out all possibilities for the minimal polynomial except $\Phi_1$, $\Phi_2$, $\Phi_1 \Phi_2$, $\Phi_3$, $\Phi_4$, and $\Phi_6$. And each of these divides one of the polynomials $x^k - 1$, for $k = 2$, $3$, $4$, or $6$. Simple examples show that all of these orders do happen.) – leslie townes Jun 14 '12 at 22:45
up vote 8 down vote accepted

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 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.

share|cite|improve this answer
I think that provides exactly the sort of impossibility proof that I was expecting might be doable by someone with (much) more background than I have in these things - many thanks! Assuming nobody produces anything incredibly startling, (and when I have had a bit of time to absorb your argument fully!) I will accept this answer. – Old John Jun 14 '12 at 22:59
I think that is brilliant - many thanks! – Old John Jun 14 '12 at 23:24

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.