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

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

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

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!

• The crux here is that finite abelian groups only have 1-dimensional irreducible representations. So your 2-dimensional rep decomposes into two 1-dim reps, which give a basis where all your matrices are diagonal. – Steve D Jun 18 '18 at 19:29
• By the way, it's more interesting to classify them up to conjugation (which turns out to coincide in this precise case). – YCor Jun 20 '18 at 22:24

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

• Here 'codiagonalizable' means that there is an invertible matrix $U$ such that $UAU^{-1}$ is a diagonal matrix for all $A$ in such finite abelian subgroup, right? – bellcircle Jun 18 '18 at 15:26
• Yes, that's exactly it (any finite set of commuting diagonalizable matrices is codiagonalizable) – Max Jun 18 '18 at 16:45

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.

• Which page is the proof written? – bellcircle Jun 18 '18 at 14:29
• Starting with page $9$, Proposition $10$. – Dietrich Burde Jun 18 '18 at 14:33
• Thanks for your answer, but the proof is too long and demanding to be a solution for qualifying exam, so I'd rather choose Max's answer below. – bellcircle Jun 18 '18 at 15:33