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

Question

$\newcommand\R{\mathbb R} \newcommand\gl{\mathrm{Gl}} \newcommand\sl{\mathrm{SL}} \newcommand\so{\mathrm{SO}}$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.

Answer 1

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.