Cohomology = invariant forms

[neworder1]

Prove the following result:

let $$G$$ be a compact Lie group, $$H$$ its closed subgroup and $$X = G/H$$. Let $$T(X)$$ denote the space of $$G$$-invariant differential forms on $$X$$ (e.g. $$\omega \in T(X) \Leftrightarrow \forall g \in G g^{*}\omega = \omega$$). Then $$T(X)$$ is isomorphic to $$H^{*}(X)$$, de Rham cohomology space of $$X$$,

Do you know where I can find the proof of this result?

I have been suggested the following proof strategy:
a) if $$\omega$$ is $$G$$-invariant, then d$$\omega = 0$$
b) likewise, d$$*\omega = 0$$ (Hodge star)
c) by Hodge theory, $$\omega$$ is harmonic, and each cohomology class has exactly one harmonic representant

Unfortuately, this is not an elementary proof. But perhaps at least a) and b) can be proved easily? A concept for proving a): locally, we can find $$G$$-invariant coordinates (i.e. a local basis of $$G$$-invariant vector fields which span the tangent space) - how to prove this? In these coordinates $$\omega$$ has constant coefficients (why?), so d$$\omega = 0$$. How about d$$*\omega$$?

I'd be glad if someone could help with filling in the details.

 

[zhentil]

Have you considered looking at it as a fibre bundle and applying the Kunneth formula to it? I must admit that I don't have too much experience with this, but that seems to be the only way to get a quick proof.