Jtdylktuy

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 $2times 2$ 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 $Bin mathfrak h$, 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.