What is known about this group reminiscent of the anharmonic group?

Question

The anharmonic group is this nonabelian group of six rational functions with the operation of composition of functions: \begin{align} t & \mapsto t & & \text{order 1} \\[8pt] t & \mapsto 1/t & & \text{order 2} \\ t & \mapsto 1-t & & \text{order 2} \\ t & \mapsto t/(t-1) & & \text{order 2} \\[8pt] t & \mapsto 1/(1-t) & & \text{order 3} \\ t & \mapsto (t-1)/t & & \text{order 3} \end{align} The reason it is called "anharmonic" appears to be that a set of four numbers is said to divide the line harmonically if their cross-ratio is $1$, and so the cross-ratio measures deviation from harmonic division, and when four numbers with cross-ratio $t$ are permuted, this group gives the six values that the cross-ratio can take. The members of this group permute the elements $0$, $1$, and $\infty$ of $\mathbb C\cup\{\infty\}$.

Today I noticed that something very similar-looking forms a group of four elements, each of the three non-identity elements having order $2$: \begin{align} t & \mapsto t \\ t & \mapsto -1/t \\ t & \mapsto (1-t)/(1+t) \\ t & \mapsto (t+1)/(t-1) \end{align}

Can anything interesting be said about this, including, but not limited to, relevance to geometry, algebra, combinatorics, probability, number theory, physics, or engineering?
What other finite groups of rational functions as simple as these exist?

The classification of finite groups of Mobius transformations is known and classical. They are classified by Dynkin diagrams. Look up "finite subgroups of SU(2)" and "McKay correspondence." — Qiaochu Yuan, Mar 12 '16 at 01:23
It seems that it's known that all finite subgroups of $\text{PGL}(2,\mathbb R)$ are cyclic or dihedral. Depending on what you mean by "as simple as these", you might glean some information from this paper: http://home.wlu.edu/~dresdeng/papers/nine.pdf. The largest group with nice integer coefficients is $D_6$. — Erick Wong, Mar 14 '16 at 22:33
@ErickWong : Maybe you should make your comment an answer. $\qquad$ — Michael Hardy, Mar 15 '16 at 03:36
@QiaochuYuan : Maybe you should make your comment an answer. $\qquad$ — Michael Hardy, Mar 15 '16 at 03:36
I will add a link to the previous question about the group of six functions: Name and role of a particular finite group? — Martin Sleziak, Oct 23 '21 at 14:00

score 4 · Accepted Answer · answered Mar 15 '16 at 21:17

The group of invertible rational functions, which explicitly consists of Mobius transformations $z \mapsto \frac{az + b}{cz + d}$, is abstractly the group $PSL_2(\mathbb{C})$. Its finite subgroups are known: by an averaging argument they correspond to finite subgroups of $PSU(2) \cong SO(3)$, the group of orientation-preserving isometries of the sphere $S^2$. There are two infinite sequences of such subgroups, the cyclic groups $C_n$ and the dihedral groups $D_n$, and then three "exceptional" groups given by the symmetry groups of the Platonic solids:

$A_4$, the symmetry group of the tetrahedron.
$S_4$, the symmetry group of the cube and the octahedron.
$A_5$, the symmetry group of the icosahedron and dodecahedron.

The subgroups you've identified are two copies of the dihedral subgroups $D_3$ and $D_2$ respectively. There are many copies of the cyclic and dihedral groups, each corresponding to rotations about a different axis.

For more on this the keyword is the McKay correspondence. These finite groups also show up in Galois theory because each of these finite groups $G$ acts on $\mathbb{C}(t)$ by automorphisms, and so we get $\mathbb{C}(t)$ as a Galois extension of $\mathbb{C}(t)^G$ with Galois group $G$.

What is known about this group reminiscent of the anharmonic group?

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