Q: Why ${\rm Lie}(H)=\{X \in {\rm Lie}(G):\exp(tX) \in H\}$? I encountered a theorem about Lie algebra of Lie subgroups. It reads (for $G$ a Lie group and $H$ a Lie subgroup) :
${\rm Lie}(H)=\{X \in {\rm Lie}(G):\exp(tX) \in H\}$.
First, I don't really understand this intuitively.
Second, when I think about it, the inclusion "$\subset$" I think I understand :
$X\in {\rm Lie}(H) \iff tX \in {\rm Lie}(H)$ since ${\rm Lie}(H)$ is a $\mathbb{R}$-vector space so by definition of the exponential mapping, we must have $\exp(tX) \in H$. But if $\exp(tX)\in H$ why must we have $X\in {\rm Lie}(H)$?

A: We can view $\operatorname{Lie}(H)=\mathfrak{h}$ as $T_eH\subseteq T_eG=\mathfrak{g}$. That is, we can view the Lie algebra as the tangent space at the identity. The tangent space consists of the trajectories of curves through $e$. That is, in general we can write for $p\in M$ a manifold
$$ T_pM=\{\gamma'(0)\:|\:\gamma:(-\varepsilon,\varepsilon)\to M\:\text{is smooth}, \gamma(0)=p\}.$$
So, in this case we can see
\begin{equation}
T_eH=\{\gamma'(0)\:|\:\gamma:(-\varepsilon,\varepsilon)\to H\:\text{is smooth}, \gamma(0)=e\}.
\end{equation}
Now, choose $X\in \mathfrak{h}$. $L(X)=\{tX:t\in\Bbb{R}\}\subseteq \mathfrak{h}$ is a one-dimensional linear subspace. Define $\gamma_X: L(X)\to H$ by sending $tX\mapsto \exp(tX)$. By the properties of the exponential,
$$ \gamma_X'(0)=\frac{d}{dt}\bigg|_{t=0}\exp(tX)=X$$
and it follows that $X\in \frak{h}$. So, every element of $\mathfrak{h}$ is the exponential of an element of $\mathfrak{g}$ (in fact of an element of $\mathfrak{h}$). Conversely, we claim that given an element of $Y\in\mathfrak{g}$ so that $\gamma(t)=\exp(tY)\in H$ for some $t\in (-\varepsilon,\varepsilon)$ and $\gamma(0)=e$, we have by the characterization above that $Y=\gamma'(0)\in \mathfrak{h}$, so the converse follows.
Intuitively, this is saying that at $e\in G,$ the tangent directions corresponding to $T_eH\subseteq T_eG$ are exactly those directions along which you can walk for a short period of time without leaving $H\subseteq G$. As a simple example, you can think of trying this with $S^1\subseteq T^2$, the torus $S^1\times S^1$.
