# What are the finite subgroups of $SU_2(C)$?

Is there any reference which classifies the finite subgroups of $SU_2(C)$ up to conjugacy ?

What I know is that lifting the finite subgroups of $SO_3(R)$ by the map $SU_2(C) \rightarrow PSU_2(C)$ gives rise to the groups : cyclic groups of even order, dicyclic groups, binary tetrahedral/octahedral/icosahedral groups. But I dont know if they are the only one.

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I claim that if you have all cyclic groups of even order, then you also have all cyclic groups of odd order! – Pete L. Clark May 20 '11 at 19:43

The image of a finite subgroup of $\text{SU}(2)$ in $\text{SO}(3)$ is a finite subgroup of $\text{SO}(3)$; moreover, the kernel is either trivial or $\{ \pm 1 \}$. But $-1$ is the unique element of order $2$ in $\text{SU}(2)$, so any group of even order contains it.

I claim all the finite subgroups of odd order are cyclic. This follows because the inclusion $G \to \text{SU}(2)$ cannot define an irreducible representation of $G$ (since otherwise $2 | |G|$), hence it must break up into a direct sum of dual $1$-dimensional representations.

So once you know the finite subgroups of $\text{SO}(3)$, you already know the finite subgroups of $\text{SU}(2)$.

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The diagonal matrix with entries 1, -1 is also unitary and of order 2, so your uniqueness claim is incorrect. – i. m. soloveichik Aug 3 '12 at 0:27
Oops-sorry--ignore my comment above – i. m. soloveichik Aug 3 '12 at 0:39

Yes, the ADE classification enumerates all possible finite subgroups of $SU(2)$. Just a trivial correction, when it comes to the Abelian cyclic groups, A, the order may be both even and odd because $U(1)$ inside $SU(2)$ has all these $Z_N$ subgroups. The odd ones aren't linked to subgroups of $SO(3)$ in the same way.

The dicyclic groups, D, and the three exceptions, E, are as you wrote.

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