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Take, for example, $f(x)=\frac{x-3}{x+1}$. One can verify that $f\circ f\circ f$ is the identity, so $f$ has order 3 in the group of Möbius transformations. Constructing such functions can be done easily.

Are there Möbius transformations of aribtrarily greater orders? If so, how can one construct them?

Fibonacci
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    From Wikipedia: "a Möbius transformation can be obtained by first performing stereographic projection from the plane to the unit two-sphere, rotating and moving the sphere to a new location and orientation in space, and then performing stereographic projection (from the new position of the sphere) to the plane." The rotations of the sphere, in particular, will contain elements of arbitrary order. – Quinn Greicius Feb 17 '16 at 06:20
  • @QuinnGreicius How to examine the rotation of the sphere? – Ooker Nov 13 '17 at 17:33
  • @Ooker Both of the current answers, including the one you accepted, are doing exactly that. – Quinn Greicius Nov 14 '17 at 04:58
  • It seems only Mobius maps corresponding to rotations of 2\pi/n (or plus mirror symmetry)have finite order – Charlie Chang Sep 02 '20 at 12:23

2 Answers2

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The composition of Möbius transforms is naturally associated with their matrix of coefficients:

$$x \rightarrow f(x)=\dfrac{ax+b}{cx+d} \ \ \ \leftrightarrow \ \ \ \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$

This correspondence is in particular a group isomorphism between the group of (invertible) homographic transforms of the real projective line and $PGL(2,\mathbb{R})$.

(composition $\circ$ mapped to matrix product $\times$).

Thus, your question boils down to the following: for a given $n$, does it exist a $2 \times 2$ matrix $A$ such that $A^n=I_2$ ?

The answer is yes for real coefficients. It suffices to take the rotation matrix :

$$\begin{bmatrix} \cos(a) & -\sin(a) \\ \sin(a) & \cos(a) \end{bmatrix} \ \ \ a=\dfrac{2\pi}{n}$$

Edit: If you are looking for integer coefficients, the answer is no. In fact, with integer coefficients, only homographies of order 2,3,4 and 6 can exist. (I rectify here an error that has been signaled and I add information). See for that the very nice paper (http://dresden.academic.wlu.edu/files/2017/08/nine.pdf) (in particular its lemma 1).

Gregory Dresden
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Jean Marie
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    The group is $PGL(2, \mathbb{C})$, not $PGL(2, \mathbb{R})$. – Michael Albanese Feb 17 '16 at 07:13
  • @Michael Albanese I don't understand what is wrong with $PGL(2, \mathbb{R})$ – Jean Marie Feb 17 '16 at 08:25
  • $a, b, c, d \in \mathbb{C}$, see here for example. – Michael Albanese Feb 17 '16 at 09:20
  • @Michael Albanese I am aware that $PGL(2, \mathbb{C})$ is the most general context, but I staid in $PGL(2, \mathbb{R})$ because it seems that the asker, implicitly, was working in the reals, and, hopefuly, as I said in my answer, in $\mathbb{Z}$. – Jean Marie Feb 17 '16 at 09:38
  • That's fine, but you said the group of Möbius transformations is isomorphic to $PGL(2, \mathbb{R})$, which is false. You're looking at a proper subgroup. – Michael Albanese Feb 17 '16 at 09:39
  • @Michael Albanese All right, I didn't understand this was the point. Would you give the group with real coefficients another name ; should I have said the group of "homographies of the real line" instead ? – Jean Marie Feb 17 '16 at 09:46
  • Let us continue this discussion in chat. – Jean Marie Feb 17 '16 at 09:47
  • Thanks, I was hoping to use an example in a calculus course I'm teaching, so obviously I can't talk about complex numbers. – Fibonacci Feb 17 '16 at 14:37
  • Also, the transformation $f(x)=\frac{x-1}{x+1}$ has order 4, as can easily be checked by hand. – Fibonacci Feb 17 '16 at 14:38
  • If you wanted to identify $PGL(2,\mathbb{R})$ as a group of "homographies", it would be more correct to call it the group of homographies of the real projective line $\mathbb{R}P^1$. Not being previously familiar with the word "homography" until I looked it up just now on Wikipedia, I would have just said "the group of projective transformations of $\mathbb{R}P^1$". – Lee Mosher Feb 17 '16 at 14:52
  • @Lee Mosher I agree. I have modified the text. – Jean Marie Feb 17 '16 at 15:23
  • I agree, Fibonacci. I made an error, writing 4 albeit I wanted to write 5. Thank you very much for spotting it. I have corrected it in the text and I have given the general result I should have given at first. – Jean Marie Feb 17 '16 at 15:29
  • How can more of those be constructed? I don't think my example is the only one. – Fibonacci Feb 17 '16 at 16:57
  • Many, many more examples can always be constructed by conjugation. For any $f,g$ in any group, if $f$ is of finite order then so is $gfg^{-1}$. – Lee Mosher Feb 17 '16 at 17:39
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Let $\zeta$ be a primitive $n^{\text{th}}$ root of unity, and consider the Möbius transformation $f(z) = \zeta z$. As

$$(\underbrace{f\circ f\circ\dots\circ f}_{k\ \text{times}})(z) = \zeta^kz,$$

the order of $f$ is $n$. Therefore, the group of Möbius transformations has an element of any finite order. In addition, $g(z) = z + 1$ provides an example of an element of infinite order.

Michael Albanese
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