Let $G$ be a (EDIT: connected) Lie group of dimension $n$, and let $\mathfrak{g}$ be the associated Lie algebra. If $x_1,\ldots,x_n$ is a basis for $\mathfrak{g},$ is it necessarily true that the 1-parameter subgroups $e^{tx_1},\ldots,e^{tx_n}$ generate $G$?
Note: It is sufficient to show that the subgroup generated by the 1-parameter subgroups is closed, since it follows that it is a Lie subgroup of dimension $n$. In particular, it must contain a neighborhood of the identity, which generates $G$.
Also, note that it is not always true that the subgroup generated by 1-parameter subgroups is a Lie subgroup; consider the 1-parameter subgroup of the 2-torus that is a line with irrational slope.
Best Answer
Let $x=\sum c_ix_i\in\mathfrak{g}$. Consider the map $f:\mathfrak{g}\to G$ given by $f(x)=e^{c_1x_1}\cdots e^{c_nx_n}$. We have $$df_0=\left.\frac{d}{dt}f(tx)\right|_{t=0}=\frac{d}{dt}(1+tc_1x_1+O(t^2))...(1+tc_nx_n+O(t^2))=\frac{d}{dt}(1+tx+O(t^2))=x.$$
This implies $df$ is surjective in a neighborhood of zero, which implies $f$ is open in a neighborhood of zero. Thus the image of $f$, which is inside the subgroup generated by the 1-parameter subgroups, contains a neighborhood of the identity, and thus is the entire group.