Could anyone help me a little with how to begin to solve the following problem? Thanks in advance!
Definitions
Let $G$ be a Lie group. For $g \in G$, define a diffeomorphism $l_g$ of $G$ by $l_g(h) = gh$. A smooth vector field $X$ is called left-invariant if $(d_hl_g)(X_h) = X_{gh}$ for all $g,h \in G$. Thus, a left-invariant vector field is completely determined by its value at the unit element $e \in G$. It is not difficult to see that every element $v \in T_eG$ arises this way: the corresponding vector field is defined by $X_g^v = (d_el_g)(v)$. Therefore the space of left-invariant vector fields can be identified with the finite dimensional space $\mathfrak g = T_eG$. The commutator of two left-invariant vector fields is again left-invariant. Hence we get a Lie bracket $[\cdot, \cdot]$ on $\mathfrak g: [v,w] = [X^v,X^w]_e$.
The problem
For $v \in \mathfrak g$, let $\gamma_v$ be the maximal integral curve of $X^v$ such that $\gamma_v(0) = e$. Show that for every $g \in G$ the curve $\gamma(t) = g\gamma_v(t)$ is an integral curve of $X^v$ such that $\gamma(0) = g$. Conclude that $\gamma_v(t)$ is defined for all $t \in \mathbb{R}$ and the flow $(\phi^v_t)_t$ defined by $X^v$ is given by $\phi^v_t(g) = g\gamma_v(t)$.
I can use the following if needed
Let $X$ be a smooth vector field on a manifold $M$. Suppose there exists $\varepsilon > 0$ such that for every $p \in M$ there is an integral curve $\gamma : (-\varepsilon, \varepsilon) \rightarrow M$ of $X$ such that $\gamma(0) = p$. Then the maximal integral curves of $X$ are defined on the whole line $\mathbb{R}$.
What I have so far
We obviously have that this is true for $e \in G$: $\gamma(t) = e\gamma_v(t) = \gamma_v(t)$. But I don't know what to do for $g \neq e$.
Anyone?
It's just a matter of unwrapping the definitions and using the fact that the tangent map $T$ is functorial:
\begin{align*}\gamma'(t) = (T\gamma)(\partial_s|_t) = T(l_g \circ \gamma_v)(\partial_s|_t) = (Tl_g \circ T\gamma_v)(\partial_s|_t) = Tl_g(X^v_{\gamma_v(t)}) = X^v_{\gamma(t)}. \end{align*}
You should be able to finish the rest yourself.