# Question about Lie bracket and right invariant vector fields

So I was thinking about the Lie bracket of right-invariant vector fields. For starters,

Let $$X, Y \in \mathcal{X}(M)$$ be smooth vector fields. The Lie bracket of $$X$$ and $$Y$$ is the vector field $$[X,Y] \in \mathcal{X}(M)$$ given by $$[X, Y] (f)= X(Y(f))-Y(X(f)), \forall f \in C^{\infty}(M).$$

It is also known that (Introduction to Smooth Manifolds by John M. Lee, Propositions 8.30 and 8.31) we have $$F_{\ast} [X,Y] = [F_{\ast} X, F_{\ast} Y]$$ whenever $$F$$ is a diffeomorphism by the naturality of the Lie bracket. Now, a right-invariant vector field is a vector field $$X$$ such that $$(R_g)_{\ast} X = X$$. The proof that the Lie bracket of two left-invariant vector fields is a left-invariant vector field goes by using that $$(L_g)_{\ast} [X,Y] = [(L_g)_{\ast} X, (L_g)_{\ast} Y].$$ By the same logic, couldn't I show that the Lie bracket of two right-invariant vector fields is a right-invariant vector field? That is, $$(R_g)_{\ast} [X,Y] = [(R_g)_{\ast} X, (R_g)_{\ast} Y].$$

The reasoning why I'm asking this is because it seems that the Lie bracket of right invariant vector fields will come with a minus sign, as shown in this answer, that doesn't appear in my derivation of the formula.