Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. The Casimir element $\Delta\in \mathfrak{zu}(\mathfrak{g})$, considered as an operator on $C^{\infty}(G)$ is right invariant, that is, $\Delta R_gf=R_g\Delta f$.
One proof of this fact (see e.g. Lang's $SL_2(\mathbb{R})$) is along these lines: $$f(x\exp(tY)g)=f(xg\exp(tg^{-1}Yg))$$ and therefore $$\Delta R_gf=R_g\text{Ad}(g)\Delta f=R_g\Delta f$$ since $\text{Ad}(g)\Delta=\Delta$.
However, I've seen another proof of the right invariance of the Casimir for compact Lie groups that I don't fully understand. The author uses the Adjoint invariance of the inner product on the compact Lie group to show that the inner product on the Lie algebra is also $dr_g$ invariant. Therefore, because of the basis-independence of the definition of Casimir, $dr_g \Delta=\Delta$.
My question is:
how to complete the proof of right invariance of Casimir given that $dr_g\Delta=\Delta$.
(The author simply says that "the result follows" but is is not clear to me how $\Delta R_g=R_g\Delta $ follows from $dr_g\Delta=\Delta$.)
Notation: $r_g$ is the right translation of the group elements by $g$, that is, $r_g(x)=xg$. And $R_g$ is the right translation of the functions by $g$, that is, $R_g(f)(x)=f(xg)$. So, $Rg(f)=f\circ r_g$.
Also, if there are other proofs of the same statement that you prefer to the ones that I have sketched, I would be thankful if you could share them here.