Finite abelian subgroups of $\mathrm{SL_2}(\mathbb{C})$

Question

This problem is from a Ph.D Qualifying Exam for algebra.

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Question: Classify all finite abelian subgroups of $\mathrm{SL_2}(\mathbb{C})$ up to isomorphism.

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My attempt: First, given a positive integer $n$, I tried to find an element of $\mathrm{SL_2}(\mathbb{C})$ of order $n$, and that was easy; a rotation by $2\pi/n$ radians. Therefore, for every $n\ge 1$, $\mathbb{Z}_n$ is a subgroup of $\mathrm{SL_2}(\mathbb{C})$.

My conjecture is that there are no other finite abelian subgroups than what I mentioned above, and I have to prove or disprove it, and here is where I stuck.

Does anyone have ideas?

Any hints or advice will help a lot!

Answer 1

Well a finite abelian subgroup has finite exponent $n$, and so all its elements are roots of $X^n-1$ and so its elements are codiagonalizable.

Now a diagonal element of $SL_2(\mathbb{C})$ is essentially an element of $\mathbb{C}^\times$; but here it has finite order. Hence your subgroup is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ (details left to the reader)

Conversely, of course any such group is a subgroup of $SL_2(\mathbb{C})$. Note that this actually yields a characterization up to conjugacy, so better than up to isomorphism

Answer 2

In his "Vorlesungen über das Ikosaeder" [Klein, 1993], published in 1884, Felix Klein gives the classification of all finite subgroups of $SL(2,\Bbb{C})$ up to conjugacy. The abelian ones are just the cyclic groups $\Bbb{Z}/n\Bbb{Z}$ for all $n\in \Bbb{N}$. For a proof see here.