Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. If $\mathfrak{h}$ is a Lie-subalgebra of $\mathfrak{g}$, then the image of $\mathfrak{h}$ under the exponential map is supposed to be a subgroup of $G$.
My question is how to prove this, preferably using the definition of the exponential map via the flow of left-invariant vector fields, or the first one from Wikipedia and without using the Baker-Campbell-Hausdorff formula.
This is not true in general. The image of $\mathfrak h$ generates a subgroup of $G$, but the image might not be the whole subgroup. For example, if $\mathfrak g$ is the set of all $2\times 2$ real matrices, and $\mathfrak h$ is the subalgebra of trace-free matrices, then the smallest subgroup of $\text{GL}(2,\mathbb R)$ containing $\exp(\mathfrak h)$ is $\text{SL}(2,\mathbb R)$; but the following matrix is in $\text{SL}(2,\mathbb R)$ but not in the image of the exponential map: $$ A= \left( \begin{matrix} -\frac12 & 0 \\ 0 & -2 \end{matrix} \right). $$ To see why, note that if $A = e^B$ for some $B\in \mathfrak h$, then $C = e^{B/2}$ would be a real matrix whose square is $A$. But a straightforward computation shows that there is no such matrix.