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Let $G$ be a connected Lie group with identity $e$ and let $g\in G$. Must there exist a left-invariant vector field $X_g = dL_g(v)$ (for some $v \in \mathfrak g$) such that the flow $\phi_t(e)$ of $X$ through $e$ passes through $g$? I believe this is equivalent to the question of whether the exponential map from $\mathfrak g$ to $G$ is surjective.

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No. The equivalence you describe holds, and it's well-known that the exponential map fails to be surjective for $\text{SL}_2(\mathbb{R})$. Any $g$ with this property necessarily has a square root, which $\left[ \begin{array}{cc} -1 & 1 \\\ 0 & -1 \end{array} \right]$ doesn't. (Since any square root in $\text{SL}_2(\mathbb{R})$ has determinant $1$, it must have eigenvalues $\pm i$, so in particular it is diagonalizable.)

The exponential map is surjective if $G$ is either compact or nilpotent.

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    Thank you @AmiteshDatta. So using Riemannian geometry, we require the compactness of $G$ to equip it with a **bi**-invariant metric and using Lie groups we require compactness to use Cartan's theorem on maximal tori.2012-06-27