Does local flow of left-invariant vector field commute with the left-translation operator? Let $G$ be a Lie group and $X$ a left-invariant vector field over $G$ (i.e. $\forall g,p\in G: (D_p l_g)(X_p) = X_{gp}$ whereby $l_g$ is the map $G\rightarrow G:p\mapsto gp$). Let $\phi_t$ be the local flow of $X$.
Is it true, that $\phi_t$ and $l_g$ commute for every $t\in \mathbb R$ and $g\in G$?
(Reason for my question: I want to solve an exercise. I have found a solution if the above statement is true. So I wonder whether from $\forall g\in G: (D_p l_g)(X_p) = X_{gp}$ follows $l_g\circ\phi_t=\phi_t\circ l_g$).
 A: In the book "Differential Manifolds" by Colon I have found the following proposition:
"[For each left-invariant vector field $X$] there is a unique 1-parameter subgroup $s_X:\mathbb R \rightarrow G$ such that $\dot s_X(0)=X_e$. Furthermore, $X$ is a complete vector field, the flow that it generates being given by $\Phi^X:\mathbb R\times G\rightarrow G$, $\Phi_t^X(a)=a\cdot s_X(t)$" [page 165, second edition]
Thereby a 1-parameter subgroup $s_X$ is a map $\mathbb R\rightarrow G$ with $s_X(0)=e$ and $s_X(t_1+t_2)=s_X(t_1)s_X(t_2)$ ($e$ is the neutral element of $G$).
Let $l_g:G\rightarrow G:p\mapsto g\cdotp$ be the left translation map for a given $g\in G$ and let $\Phi^X$ be the global flow of $X$ as in the above propostion. We have
$$ (l_g\circ \Phi^X_t)(a)=l_g(\Phi^X_t(a))=l_g(a\cdot s_X(t))=g\cdot a\cdot s_X(t) $$
and
$$ (\Phi^X_t\circ l_g)(a) = \Phi^X_t(l_g(a))=\Phi^X_t(g\cdot a)=g\cdot a\cdot s_X(t) $$
Thus the left translation map $l_g$ and the global flow $\Phi^X_t$ commute for every $t\in \mathbb R$ and $g\in G$.
