# Proving that SO(3) is not isomorphic to another group

$$\newcommand\R{\mathbb R} \newcommand\gl{\mathrm{Gl}} \newcommand\sl{\mathrm{SL}} \newcommand\so{\mathrm{SO}}$$How do you prove that $$\gl(2,\R)/\R^*$$ is not isomorphic (as abstract groups) to $$\so(3,\R)$$? Initially I thought that because $$\gl(2,\R)/\R^*$$ is isomorphic to $$\sl(2,\R)/\{-I,I\}$$ then it is easier to look at $$\sl(2,\R)$$ but I am not sure anymore. Anything remarkable to know about centralizers and normalizers that could help me here? Actually I am no event sure anymore if they are not isomorphic.

• To see that they are not isomorphic as topological groups you can show that one is compact and the other one is not. However, this itself does not exclude the possibility that there is a group isomorphism that is not a homeomorphism, so you're probably looking for a further argument. Feb 3, 2022 at 8:34
• I'm still glad there is a topological argument showing they are not homeomorphic as topological groups. Any argument in the category of groups is also welcome of course. Feb 3, 2022 at 8:39
• You should specify if you are looking for an isomorphism of abstract groups or of Lie groups. The group $SO(3)$ is simple (as an abstract group), while $PGL(2,R)$ is not. (Why?) Another argument will tell apart $SO(3)$ from the (abstractly) simple group $PSL(2,R)$: Consider finite subgroups that they contain. Could you spot the difference? Feb 3, 2022 at 8:41
• @Moishe Kohan: Thanks! Apologies for not clarifying (I just made an edit that did so). Could you write this same answer in a new answer post so I could mark it as answered. Feb 3, 2022 at 9:39

1. The group $$SO(3)$$ is simple as an abstract group, see e.g. here. On the other hand, $$PGL(2, {\mathbb R})$$ contains $$PSL(2, {\mathbb R})$$ as (an/the) index 2 subgroup, hence, is not simple.
2. Another argument is to consider finite subgroups. The group $$SO(3)$$ contains the simple subgroup $$A_5$$ as the group of orientation-preserving symmetries of regular dedecahedron. On the other hand, each finite subgroup of $$PGL(2, {\mathbb R})$$ is conjugate into its maximal compact subgroup $$O(2)$$ and, hence, is either dihedral or cyclic. This argument also tells apart $$PSL(2, {\mathbb R})$$ from $$SO(3)$$.
3. Yet another, harder, argument is to consider infinite solvable subgroups: $$PGL(2, {\mathbb R})$$ contains the class 2 solvable subgroup of affine transformations of the real line. In contrast, one can show that every solvable subgroup of $$SO(3)$$ contains an abelian subgroup of index at most 2.