= Solution
Choose a basis $T_a$ of the <Lie algebra>, with $[T_b,T_c]=c^a{}_{bc}T_a$. On the <Matrix Lie group> set
$$
\lambda=g^{-1}dg=T_a\lambda^a,\qquad \rho=dg\,g^{-1}=T_a\rho^a.
$$
Left multiplication by a constant matrix preserves $\lambda$, while right multiplication preserves $\rho$. Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Apply the <graded Leibniz rule> to obtain
$$
d\lambda=-\lambda\wedge\lambda,
\qquad d\rho=+\rho\wedge\rho.
$$
Because the <wedge product> antisymmetrizes the matrix products, these <Maurer-Cartan equations> become
$$
\boxed{d\lambda^a=-\tfrac12c^a{}_{bc}\lambda^b\wedge\lambda^c,
\qquad d\rho^a=+\tfrac12c^a{}_{bc}\rho^b\wedge\rho^c.}
$$
Follow the vector-field naming used in the question: $L_a(g)=T_ag$ is right-invariant and generates left multiplication, whereas $R_a(g)=gT_a$ is left-invariant and generates right multiplication. They are dual to $\rho^a$ and $\lambda^a$, respectively. For any <one-form> $\eta$, $d\eta(X,Y)=X(\eta(Y))-Y(\eta(X))-\eta([X,Y])$. Evaluating the two <Maurer-Cartan equations> on the corresponding invariant fields yields
$$
\boxed{[L_a,L_b]=-c^c{}_{ab}L_c,\qquad[R_a,R_b]=c^c{}_{ab}R_c,\qquad[L_a,R_b]=0.}
$$
For the last identity, the left and right multiplication flows commute. The opposite sign for the <right-invariant vector fields> is therefore intrinsic; the letters attached to the two families are only a convention.
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