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The group generated by the functions $x\mapsto 1/x$ and $x\mapsto 1-x$ with composition of functions as the group operation is a non-abelian group with only six elements (listed below).

Does this particular realization of $S_3$ have a conventional name? Is there a literature concerning it? Does it interface in interesting ways with geometry, probability, combinatorics, algebra, number theory,${}\,\ldots\,{}$?

\begin{array}{rcl|c|cccc} & & & \text{order} & \text{fixed points} \\ \hline x & \mapsto & x & 1 & \mathbb C\cup\{\infty\} \\[10pt] x & \mapsto & 1/x & 2 & \pm 1 \\[6pt] x & \mapsto & 1-x & 2 & 1/2,\ \infty \\[6pt] x & \mapsto & x/(x-1) & 2 & 2,\ \infty \\[10pt] x & \mapsto & (x-1)/x & 3 & (1\pm i\sqrt 3)/2 \\[6pt] x & \mapsto & 1/(1-x) & 3 & (1\pm i\sqrt 3)/2 \end{array}

PS: Since this is isomorphic to the group of permutations of three elements, one could wonder which three elements it permutes, and it seems we should take those to be $0$, $1$, and $\infty$.

The first element of order $2$ transposes $0$ and $\infty$ while leaving $1$ fixed. The second transposes $0$ and $1$ while leaving $\infty$ fixed. The third transposes $1$ and $\infty$ while leaving $0$ fixed.

The first element of order $3$ takes $0$ to $\infty$, $\infty$ to $1$, and $1$ to $0$. The second takes $0$ to $1$, $1$ to $\infty$, and $\infty$ to $0$.

PPS: Evaluate the derivative of each of these at one of the fixed points. Unsurprisingly, you get a primitive $n$th root of $1$, where $n\in\{1,2,3\}$ is the order. The fixed point is the axis of a rotation of $0^\circ$, $180^\circ$, or $120^\circ$.

Michael Hardy
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  • Well, it is $S_3$ as that is the only non-abelian group with $6$ elements. But I assume you mean in this "disguise"? – Tobias Kildetoft Jul 06 '15 at 19:57
  • In general, the group generated by some finite set of involutions can be thought of as a Coxeter group – Ben Grossmann Jul 06 '15 at 19:58
  • @Omnomnomnom In this case, yes. In general, only if the relations are of the form $(st)^m$ for various $m$ and simple generators $s$ and $t$. – Tobias Kildetoft Jul 06 '15 at 20:01
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    @TobiasKildetoft : Of course it is $S_3$, but it is also a set of rational functions, and things can be said about rational functions. – Michael Hardy Jul 06 '15 at 20:01
  • @MichaelHardy I might recommend editing the question to emphasize you mean this group specifically with those elements as a set, because, in questions like this, it's usually tacitly assumed one means "the isomorphism class of this group," not a very specific incarnation. – Adam Hughes Jul 06 '15 at 20:09
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    As you seem to have found out, it's $\operatorname{Aut}(\widehat{\mathbb{C}}\setminus {0,1,\infty})$, the standard representative of the thrice-punctured sphere. – Daniel Fischer Jul 06 '15 at 20:53
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    I disagree with the isomorphist critics. That viewpoint is one viewpoint one can take towards groups, and a good one to take when doing pure group theory. It shouldn't be an ideology that prevents us from talking about and naming sets with group structure in their respective contexts. In some contexts it's absolutely crucial to distinguish different isomorphic groups -- you wouldn't complain about treating the rotations about the $x$ axis as distinct from the rotations about the $y$ axis. At least I hope you wouldn't... – joriki Jul 06 '15 at 21:38
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    First time I've ever seen the word "isomorphist". ${}\qquad{}$ – Michael Hardy Jul 06 '15 at 21:58

2 Answers2

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It is (of course?) a particular subgroup of the group of Möbius transformations and, as such, of $PGL(2,\mathbb{C})$; apparently in this guise it's specifically known as the anharmonic group and you may have some luck looking for that name.

Steven Stadnicki
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  • Do know the motivation for the term anharmonic here? – Travis Willse Jul 07 '15 at 03:54
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    @Travis : When I was 15 I read C. Stanley Ogilvy's Excursions in Geometry, an excellent book. In it I read about harmonic division. Four points divide a line harmonically iff their cross-ratio is $1$. Thus the cross-ratio measures how far a quadruple of points is from dividing the line harmonically. Hence it is called the anharmonic ratio. The cross-ratio is a function of four arguments, hence there are $24$ permutations of them. But they come in sets of four: for each of the four, the cross-ratio is the same. https://en.wikipedia.org/wiki/Harmonic_division – Michael Hardy Jul 07 '15 at 05:29
  • . . . . Hence there are size possible values of the cross-ratio, given the four points. I've actually seen this before, but it hadn't occurred to me when I posted the question. – Michael Hardy Jul 07 '15 at 05:30
  • @MichaelHardy Many thanks. I'd never seen that (particularly nice) interpretation of the cross-ratio before. – Travis Willse Jul 07 '15 at 05:37
  • I meant: "Hence there are six possible${},\ldots$" $\qquad{}$ – Michael Hardy Jul 07 '15 at 05:51
  • I've just done some editing of the following section, which is still less than perfect: Six cross-ratios and the anharmonic group ${}\qquad{}$ – Michael Hardy Jul 07 '15 at 18:21
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This particular representation of $S_3$ appears in the theory of elliptic curves, specifying the fractional linear transformations that can be applied to a parameter $\lambda \in \mathbb{C} \setminus \{ 0, 1 \}$ that do not change the isomorphism class of the elliptic curve (in Legendre normal form).

$$y^2 = x(x - 1)(x - \lambda).$$

See this blog post for some details. The value of $\lambda$ itself, separate from its orbit under this action, describes an elliptic curve with level structure where $N = 2$. The orbit of $\lambda$ under the action also describes the possible cross ratios of the four $2$-torsion points of the elliptic curve, namely $0, 1, \infty$, and $\lambda$.

Qiaochu Yuan
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