Please assume that $G/S$ is an abelian scheme if it helps.

Let $G/S$ be a commutative group scheme and let $(\Omega^\bullet_{G/S},d)$ be the algebraic de Rham complex. Let $\omega\in \Omega^r_{G/S}(G)$ be an invariant r-form. Is it true in this generality that $d\omega=0$? Is there a reference for such a fact?

By ulrich's answer below the statement is true whenever 2 is invertible on $G$. But what if $G$ has points of characteristic 2?

I'm trying to understand the proof of the degeneracy of the Hodge-de Rham spectral sequence for abelian schemes from Berthelot-Breen-Messing, Théorie de Dieudonné cristalline II, specifically Lemma 2.5.3. In its proof they say "Comme les différentielles invariantes sont fermées[...]".

Lie Groups and Lie Algebras, which is stated in terms of Lie groups and Lie algebras, of course---possibly over a field of characteristic > 0---but which also applies to smooth algebraic groups. (Bourbaki's formulation avoids the "1/2" that may have worried you in other treatments.) The starting point is the familiar formula d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X,Y]): if X,Y, and \omega are left invariant, then the first two terms vanish... $\endgroup$ – user2490 May 30 '11 at 4:13