how to find an integral curve in Lie group? Given a Lie group $G$, $e$ is its identity element and $g$ is one element of $G$. I want do find a curve $\gamma(t)$ that satisfies these conditions:
1) passes $g$ and $e$, that is $\gamma(0)=e,\gamma(t_g)=g$;
2) the tagent vector at $e$ is $v\in T_eG$, that is $\gamma'(t)|_{t=0}=\xi$.
Does this curve exist in a left invariant vector field? If yes, how to find it? If no, why? can we find it in other type of vector field?
My try: let $\gamma(t)=\exp(\xi t)$, so $\gamma(0)=e$, then solve $\gamma(t_g)=\exp(\xi t_g)=g$, get $t_g$, then we have the vector at $g$: $\xi_g=\gamma'(t)|_{t=t_g}$. I dont think my solution is right, but I dont know where is the problem, and how to get a better solution.
Thank you.

In general no for left invariant vector fields, since $\gamma(t_g)=\exp(t_gv)$ may not be equal to the $g$ you want. For general case, of course...say $g=\exp(t_gu)$, then define $\gamma(t)=\exp(tv+t^2(u-v)/t_g)$.
