How to find all abelian subgroups of Möbius transformations?
It's a problem from conway's functions of one complex variable.
The things I have known is that when two Möbius transformations commute,they will have the same fixed point.
And I think this is related to abelian groups of matrix group since the composition of transformations is just like matrix multiplication.
Any help will be thanked.
Mobius transformations come from $PSL_2(\mathbb{C})$, which act sharply $3$-transitively on the Riemann sphere. (That is, given three distinct complex numbers $w_1,w_2,w_3$ and three others $z_1,z_2,z_3$, including $\infty$ as a complex number, there is exactly one $g\in PSL_2(\mathbb{C})$ for which $z_i=gw_i$ for $i=1,2,3$.)
Let's classify abelian subgroups $A\le PSL_2(\mathbb{C})$ up to conjugacy.
If $|A|=2$ then the nontrivial element of $A$ is an involution $\alpha$. If $\alpha$ is nontrivial it must contain a $2$-cycle (as a permutation). We may conjugate $A$ so this $2$-cycle is $(0\infty)$. Let $\alpha=[\begin{smallmatrix} a & b \\ c & d\end{smallmatrix}]$; I leave it as an exercise to check $\alpha(0)=\infty$ and $\alpha(\infty)=0$ imply $\alpha=[\begin{smallmatrix} 0 & -\lambda \\ \lambda^{-1} & 0 \end{smallmatrix}]$ which is conjugate to $[\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}]$ by $[\begin{smallmatrix} \sqrt{\lambda} & 0 \\ 0 & \sqrt{\lambda}^{-1} \end{smallmatrix}]$.
Otherwise if $|A|>2$ then suppose $g,h\in PSL_2(\mathbb{C})$ are nontrivial. You say they must share the same fixed point; by conjugating $A$ we can ensure this fixed point is $\infty$ so both $g$ and $h$ are upper triangular and the Mobius transformations are just affine transformations of the form $g(z)=az+b$.
Case $a=1$. Then $g$ is unitriangular. I leave it as an exercise to check $h$ must also be represented by a unitriangular matrix, so $A$ is a subgroup of the unitriangular matrices.
Case $a\ne 1$. Note $b/(1-a)$ is another fixed point of $az+b$ (by simple calculation); conjugating $A$ we can ensure this other fixed point is $0$, so then $g$ is diagonal. I leave it as an exercise to check this means $h$ must also now be diagonal, so $A$ is any subgroup of the diagonal matrices represented in $PSL_2(\mathbb{C})$.