# Discrete subgroups of SU(n) and SO(n).

Thank you very much for your concern. I am in physics background, any simpler but complete explanation would be helpful.

I would like to know whether there is a complete understanding of discrete subgroup of SU(n) and SO(n). If SU(n) and SO(n) are too general, in particular, we can focus on:

(1) all discrete subgroups of SU(2),

(2) all discrete subgroups of SU(3),

(3) all discrete subgroups of SO(3),

(4) all discrete subgroups of SO(4).

I have heard the fact of discrete subgroups of SU(2) is addressed by ADE classification. I also learned from a note about discrete subgroups of SO(3) by Ginzparg's note. But I cannot list down all of the subgroups of SU(2) and SO(3), for example:

(5) whether $(\mathbb Z_2)^3$ or $\mathbb Z_n \times \mathbb Z_m \times \mathbb Z_o$ (with some arbitrary $n,m,o$) are subgroups of SU(2), SU(3), SO(3), SO(4)?

I would be happy to know whether the facts about these groups, such as:

SU$(2)/\mathbb Z_2=$ SO(3)

SO$(4)/\mathbb Z_2=$ SO(3) $\times$ SO(3)

SO$(4) =($SU(2) $\times$ SU(2))/$\mathbb Z_2$

whether these facts can help address the subgroups of SU(2), SU(3), SO(3), SO(4) from one to the other, i.e. if I understand all discrete subgroups of SO(3), can I obtain all discrete subgroups of SU(2) from SU$(2)/\mathbb Z_2=$ SO(3)?

PS. Accidentally I found this paper by Hanany and He, where its Appendix I and III offer table for discrete subgroups of SU(2) and SU(3); however, further illumination on discrete subgroup of SU(n) and SO(n) and my (1), (2), (3), (4), (5) would still be helpful.

See en.wikipedia.org/wiki/Category:Binary_polyhedral_groups for finite subgroups of $\text{SU}(2)$, which double cover the finite subgroups of $\text{SO}(3)$. It also turns out that $\text{SU}(2) \times \text{SU}(2)$ double covers $\text{SO}(4)$ (but they are not isomorphic as you claim), so the finite subgroups of the latter can be understood in terms of the finite subgroups of the former; this classification is somewhere in Conway and Smith's On Quaternions and Octonions. –  Qiaochu Yuan May 21 at 1:22
Note that every finite group is a subgroup of the symmetric group on $n$ letters for some $n$, and so occurs as a discrete subgroup of $O(n)\subset SO(n+1)$, so the complete understanding of discrete subgroups of $SO(N)$ should be pretty hard. –  Dalimil Mazáč May 21 at 16:13