I want to solve the following question
Show that if a non-trivial element of $\mathcal{M}$ has finite order, then it fixes precisely two points in $\mathbb{C}_{\infty}$. Hence show that any finite abelian subgroup of $\mathcal{M}$ is either cyclic or isomorphic to $C_{2} \times C_{2}$.
My Proof: I managed to do the first part by using the fact that there are three conjugacy classes, meaning that either all,1 or 2 elements are fixed. Since we are dealing with an non identity element, it follows that either 1 or 2 elements are fixed. Then writing out the mobius transforms in matrix forms in some convenient basis and abusing the fact that the order is finite, we get that the only possibility is that we have only two fixed points.
The second part of the question is where I am having issues. I was able to prove that if $f,g \in H$, where $H$ is the subgroup in question, then $f$ and $g$ fix the same two elements, or interchanges them (just by looking at the fact that it is abelian). Using the fact that it has finite order I was also able to prove that each element has order $2$. But here is where I am stuck. I know that I have to show that for any other $h \in H$, we have that
$$
h \in \langle f,g \rangle
$$
but I can't seem to prove that. I tried
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Using the abelian property to show that $fh=hf$ and $gh=hg$ fix what $h$ can be and then go by contradiction with the size of the group and Lagrange's theorem, but to no success
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I tried using matrices and matching coefficients but that again did not work.
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I tried using cosets, by noting that $H$ must have an order that is a power of $2$ but again to no success.
Any ideas as how to proceed?
Edit + Message to the moderators: My question is not the same as the other one as the techniques and method used in the other question have not been covered in my course. This is a question from a first year course on group theory from the University of Cambridge.
Best Answer
Let $A$ be a non-trivial finite abelian subgroup of the Mobius group. Let $f\in A$. You have proved that $f$ has exactly two fixed points. As the Mobius group is $3$-transitive on the projective line $\mathbb{C}\cup\{\infty\}$ we may assume (by conjugating by a suitable element of the Mobius group) that the fixed points of $f$ are $0$ and $\infty$.
Now any element of the Mobius group fixing exactly $0$ and $\infty$ has the form $z\mapsto \lambda z$. If all the non-trivial elements of $A$ fix exactly these two points then each is of the form $z\mapsto \lambda z$ where $\lambda$ is by the finiteness a $k$-th root of unity for some $k$, so $A$ is isomorphic to a subgroup of the roots of unity in $\mathbb{C}$ and so clearly cyclic.
So suppose now that we have $f\in A$ of the form $z\mapsto \lambda z$, and also some $g$ which does not have fixed points $0$ and $\infty$. You have shown that $g$ swaps $0$ and $\infty$, and so $g$ is of the form $z\mapsto \mu/z$. Using the fact that $fg=gf$ we have that $\lambda=-1$. The same will be true for any other non-trivial element of $A$ with fixed points $0$ and $\infty$; it will be $z\mapsto\lambda' z$ and then $\lambda'=-1$.
Suppose that $g'$ is another element of $A$ swapping $0$ and $\infty$. It will be of the form $z\mapsto\mu'/z$. Using the fact that $gg'=g'g$ we find that $\mu'=\pm\mu$.
Hence if $A$ is not cyclic it consists of $\{z\mapsto z, z\mapsto -z, z\mapsto\mu/z, z\mapsto-\mu/z\}$, a copy of the fours-group $C_2\times C_2$.