 Solution

Take the frame dual to the left-invariant coframe of the translation-dilation group. It isIndeed ${\lambda }^i(L_j)={\delta }_j^i$. These are left-invariant vector fields, either because they are dual to invariant forms or because $L_j(g)=(d{L}_g{)}_e{X}_j$. For instance, the curve $g\exp (t{X}_0)$ differentiates to $\rho {\partial }_{\rho }$, while $g\exp (t{X}_1)$ differentiates to $\rho {\partial }_{x^1}$.

The coordinate Lie brackets satisfyFor the first identity only the derivative of the coefficient $\rho$ contributes. The linear map $\boxed{X_j⟼L_j}$ is a Lie algebra isomorphism: it preserves all basis brackets, is injective by independence of the frame, and is onto the three-dimensional space of left-invariant vector fields.