Let $M$ be a differential manifold. Let $\mathfrak{X}(M)$ be the set of smooth vector field over $M$. Let $g$ be a finite Lie subalgebra of $\mathfrak{X}(M)$. Let $G$ be a connected, simply connected Lie group whose Lie algebra is $g$.
Suppose $g$ has the property that $\forall X\in g$, $X$ is complete. We can thus define the transformation $X_t\in\mathrm{Diff}(M)$. The group generated by all such transformations is a subgroup of $\mathrm{Diff}(M)$, denoted $H$.
If $\dim g=1$, $\exp(tX)\rightarrow X_t$ defines a homomorphism $G\rightarrow H$.
In general, how do we prove that there exists such a homomorphism?
(in Kobayashi, page 13, it is stated that the group $G$ acts locally on $M$, which, I believe, is equivalent to the existence of such a homomorphism. The homomorphism is also used in the proof of lemma 1 on the same page).
Here I give a proof idea:
Let $N=G\times M$ and $g_\times$ be the vector field defined by $(g,m)\mapsto(X_{\mathfrak{X}(G)}(g),X_{\mathfrak{X}(M)}(m))$. Then by Frobenius theorem there exists a unique submanifold $\mathcal{L}_m$ passing by $(e,m)$ integrating $g_\times$. We prove that $\mathcal{L}_m\simeq G$ by the first projection, and there exists $\mathrm{Diff}(\mathcal{L}_m)\rightarrow\mathrm{Diff}(M)$ a group homomorphism induced by the second projection (which concerns picking a pre-image in $\mathcal{L}_m$ the choice of which which leaves the overall operation invariant). And finally we observe that the composite morphism:
$$G\rightarrow\mathrm{Diff}(G)\rightarrow\mathrm{Diff}(\mathcal{L}_m)\rightarrow\mathrm{Diff}(M)$$
sends $\exp(tX)$ to $X_t$.