# Infinite subgroups of SO(3)

The classification of finite subgroups of SO(3) is well-known: we have the cyclic groups, the dihedral groups, and the symmetries of the Platonic solids. Is there an analogous result for the infinite subgroups of SO(3)?

I don't really need a full classification, I'm interested only in subgroups that are neither Abelian nor dense in SO(3).

So far the only such subgroup I managed to find is the infinite analogue of the dihedral groups; pretty much anything you try is dense in SO(3).

• The discrete subgroups are finite. Otherwise, consider the closure of a subgroup - what can it be? I think all you get are infinite subgroups of the copy of O(2) not in SO(2).
– anon
Commented Jun 16, 2021 at 20:55

You have essentially found all the examples: every infinite non-dense subgroup of $$SO(3)$$ is contained in an infinite analogue of a dihedral group. To be precise, given a $$2$$-dimensional linear subspace $$V\subset\mathbb{R}^3$$, there is a subgroup of $$SO(3)$$ consisting of rotations that map $$V$$ to itself. This subgroup can naturally be identified with the orthogonal group $$O(V)\cong O(2)$$, since every orthogonal map $$V\to V$$ can be uniquely extended to an orthogonal map $$\mathbb{R}^3\to\mathbb{R}^3$$ of determinant $$1$$ (if $$v\in V^\perp$$ then the extension maps $$v$$ to either $$v$$ or $$-v$$ as needed to make the determinant have the correct sign). This can be thought as a sort of infinite dihedral group, being a semidirect product of the rotation group $$SO(2)$$ and a reflection $$\mathbb{Z}/(2)$$.
The claim is then that every infinite non-dense subgroup $$G$$ of $$SO(3)$$ is contained the subgroup $$O(V)$$ for some plane $$V$$. Replacing $$G$$ with its closure, it suffices to consider the case where $$G$$ is a closed proper subgroup of $$SO(3)$$. This means that $$G$$ is a Lie subgroup of $$SO(3)$$, and since $$G$$ is infinite and compact, it must have positive dimension. The connected Lie subgroups of $$SO(3)$$ can be classified using subalgebras of the Lie algebra $$\mathfrak{so}(3)$$ (see here for instance); the only nontrivial proper connected Lie subgroups are the 1-parameter rotation groups $$SO(V)$$ in planes $$V\subset\mathbb{R}^3$$. So the connected component of the identity in $$G$$ must be of the form $$SO(V)$$. If $$g\in G$$, then $$g$$ maps $$V$$ to some other plane $$V'$$ and then $$G\supseteq gSO(V)g^{-1}=SO(V')$$. This implies $$V'=V$$, since if $$G$$ contained $$SO(V')$$ for some other $$V'$$ then the connected component of the identity in $$G$$ would be larger than just $$SO(V)$$. Thus every element of $$G$$ maps $$V$$ to itself, i.e. $$G\subseteq O(V)$$.
As for classifying infinite nonabelian subgroups of $$O(V)$$, every such subgroup must contain a reflection, and so up to conjugation you can assume it contains the generator of $$\mathbb{Z}/(2)$$ in the semidirect product representation $$O(V)\cong SO(V)\rtimes \mathbb{Z}/(2)$$. It then follows that the subgroup must be $$H\rtimes\mathbb{Z}/(2)$$ for some subgroup $$H\subset SO(V)$$. So the infinite non-dense nonabelian subgroups of $$SO(3)$$ are exactly the semidirect products of an infinite group of rotations in a plane $$V$$ together with a reflection of that same plane (which from a 3-dimensional perspective is actually a 180-degree rotation around an axis contained in $$V$$).