Do local flows of left-invariant vector fields satisfy $\Phi_X^t\circ L_x=L_x\circ \Phi_X^t$? Given $G$ a lie group and $X$ a left-invariant vector field. Let $\Phi_X^t$ be the local flow of $X$. Why can we conclude that $\Phi_X^t \circ L_x=L_x \circ \Phi_X^t$?
Thanks!
 A: There is a bijection between $T_eG$, left-invariant vector fields and one-parameter subgroups. We'll be interested in correspondence between second and third thing. Let's denote by $\gamma : \mathbb{R} \to G$ one-parameter subgroup corresponding to $X$. It is locally a solution to a problem $\gamma(0) = e$, $\dot \gamma (t) = X_{\gamma(t)}$, namely: $\gamma(t) = \Phi_X^t(e)$. Let's take any $g \in G$ and define $\gamma_g (t) = g \cdot \gamma(t) = L_g(\gamma(t))$. One can observe that after applying chain rule and left-invariance of $X$ we may obtain $\dot \gamma_g (t) = D_{\gamma(t)}L_g(\dot \gamma(t)) = D_{\gamma(t)}L_g(X_{\gamma(t)}) = X_{g \cdot \gamma(t)} = X_{\gamma_g (t)}$ hence $\gamma_g$ is locally a solution to a problem $\gamma_g(0) = g$, $\dot \gamma_g (t) = X_{\gamma_g (t)}$ and hence $\Phi_X^t(g) = \gamma_g (t) = g \cdot \gamma(t)$.
$(L_x \circ \Phi_X^t)(g) = L_x(\Phi_X^t(g)) = L_x (g \cdot \gamma(t)) = x \cdot g \cdot \gamma(t) = (x \cdot g) \cdot \gamma(t) = \gamma_{x \cdot g}(t) = \Phi_X^t(x \cdot g) = \Phi_X^t(L_x(g)) = (\Phi_X^t \circ L_X) (g).$
