@johncarlosbaez I didn’t consider that! All the components of a Lie group are diffeomorphic. I forgot! So if (G) is compact and not necessarily connected, then its entire de Rham cohomology can be computed from its Lie algebra cohomology. As you said, if (G) has (n) components, then it’s de Rham cohomology is just (n) copies of (H^\bullet(\mathfrak{g}))! So the whole thing is known! Thanks!