# $S_4$ subgroups and $SO_3(\mathbb{R})$

Let $$\Gamma$$ and $$\Gamma'$$ be two octahedral subgroups (isomorphic to the 24 element group $$S_4$$) of $$SO_3(\mathbb{R})$$. Then any isomorphism $$\phi: \Gamma \to \Gamma'$$ extends to a unique automorphism of $$G=SO_3(\mathbb{R})$$. To see this note that all octahedral subgroups of $$G$$ are conjugate (indeed this is true for all isomorphic finite subgroups of $$G$$). So there always exists some $$g \in G$$ such that $$g \Gamma g^{-1}=\Gamma'$$ giving us the correct image and so agreeing with $$\phi$$ up to some automorphism. From there we can conjugate by some $$x \in \Gamma \cong Aut(\Gamma)$$ to get an exact match. So that conjugation by $$xg$$ is an automorphism of $$G$$ extending $$\phi$$.

From there we can get uniqueness by noting that every automorphism of of $$G$$ is inner and that if $$g_1,g_2$$ both conjugate $$\Gamma$$ to $$\Gamma'$$ through exactly the action of $$\phi$$ then conjugation by $$g_1g_2^{-1}$$ fixes $$\Gamma$$ pointwise. So $$g_1g_2^{-1} \in N_G(\Gamma)$$ but $$\Gamma$$ is self-normalizing $$N_G(\Gamma)=\Gamma$$ so $$g_1g_2^{-1} \in \Gamma$$. However $$\Gamma$$ also has trivial center $$Z(\Gamma)=1$$ so we must have $$g_1=g_2$$ and thus uniqueness is proven.

Let $$G$$ be a compact connected Lie group and $$\phi:S_4 \to G$$ an injective group homomorphism. What can we say about extensions $$\tilde{\phi} : SO_3(\mathbb{R}) \to G$$? Do they always exist? Are they always unique?