Lie group question: If $\gamma^{-1}\dot{\gamma}\in\mathfrak{g}$ everywhere, does $\gamma(t)\in G$? Let $G$ be a Lie subgroup of $GL(n,\Bbb R)$ and $\mathfrak{g}\subseteq M(n,\Bbb R)$ its Lie algebra. Suppose that we have a smooth curve
$$\gamma:\Bbb R\to G$$
with $\gamma(0)=I$. Then, it induces a curve on the Lie algebra
$$\alpha:\Bbb R\to \mathfrak{g},\quad \alpha(t)=\gamma(t)^{-1}\frac{d\gamma}{dt}(t).$$
I am wondering about the converse of this fact:

Question: Let $\gamma:\Bbb R\to M(n,\Bbb R)$ be a smooth curve with $\gamma(0)=I$ and suppose that
  $$\gamma(t)^{-1}\frac{d\gamma}{dt}(t)\in\mathfrak{g},\quad\forall t\in\Bbb R.$$
  Does $\gamma(t)\in G$ for all $t\in\Bbb R$?

If true, it would have the following implication:
Let $\alpha:\Bbb R\to\mathfrak{g}$ be a smooth curve and consider the initial value problem
$$\frac{d\gamma}{dt}(t)=\gamma(t)\alpha(t),\quad \gamma(0)=I.$$
Since it is a first order linear system of ODEs, it has a unique solution
$$\gamma: \Bbb R\to M(n,\Bbb R),$$
and in fact $\gamma(t)\in G$ for all $t\in\Bbb R$.
 A: Yes. Actually, it's a much more general fact. Suppose I have a smooth manifold $M$ and a foliation on it: that is, a subbundle of the tangent bundle such that, at every point, there is locally a submanifold such that the subbundle is its tangent bundle. Then I can integrate the foliation to get a partition of my manifold into disjoint immersed submanifolds (which are also weakly embedded, if you know the term).
Then given a Lie subgroup of a Lie group $H$ (Lie subgroup meaning injective immersive homomorphism $G \to H$ from another Lie group), the tangent bundles of its cosets give a foliation of $H$. (The immersed submanifolds here are the cosets.)
I'm going to ignore the above formalism now, since it doesn't help you see the trees. If you like, you can see how the following argument ports over in general.
1) That $\gamma^{-1} \dot \gamma \in \mathfrak g$ is precisely the same thing as saying that $\gamma$ is tangent to the cosets of $G$ (precisely, $\gamma'(t)$ is tangent to the coset through $\gamma(t)$). We will show that, if $\gamma(t) \in G$, so is $\gamma(t+h)$, $h$ a small number (positive or negative). So around $\gamma(t)$, use the implicit function theorem to get a chart $U$ in which the inclusion of $G$ is just the map $\Bbb R^n \hookrightarrow \Bbb R^{n+k}$, and the inclusion of $hG$ for $h$ near $1$ is $\Bbb R^n \times {c} \hookrightarrow \Bbb R^{n+k}$. In this chart, your claim is just that if $\gamma(t) \in \Bbb R^n$, and $\dot \gamma \in \Bbb R^n$ (here we are using the special form of the chart, where orbits look like parallel copies of $\Bbb R^n$), then for all times $\gamma$ remains in the chart, $\gamma$ is in $\Bbb R^n$. But this is obvious - you can just integrate! This proves the desired theorem. 
So the times for which $\gamma \in G$ forms an open set. But the times for which $\gamma \in hG$ (that is, for which $\gamma$ is in some specified coset) are also open by the exact same argument. So you've obtained a partition of $\Bbb R$ into disjoint open sets. This is only possible if only one of the open sets is nonempty; and $\gamma(0) \in G$, so $\gamma(t) \in G$ for all $t$.
