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).

  • $\begingroup$ 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). $\endgroup$
    – anon
    Commented Jun 16, 2021 at 20:55

1 Answer 1


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 $V$).


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