This is probably a trivial question, but I am not able to see how to go about it:
Let $G$ is a Lie group and let $\gamma : I \subset \mathbb{R} \rightarrow G$ is a smooth curve in $G$. Is it true that $$ \gamma' = X\gamma $$ for some $X \in \mathfrak{g}$ (where $\mathfrak{g}$ is the Lie algebra of $G$)?
I can see that I can assume without loss of generality that $\gamma(0) = e$, the group identity, and the result is true if $\gamma(t) = e^{tX}$ with $X = \frac{d}{dt}\gamma(t)|_{t=0}$.
How do I argue for an arbitrary curve $\gamma$?
It is not true.
Because the curve corresponding to an element in $\mathfrak{g}$ is a subgroup of $G$, while an arbitrary curve doesn't satisfy this condition.
For example, a curve in $SL(2,\mathbb{R})$ is given by \begin{equation} \gamma(t)=\begin{pmatrix} e^t&t\\0&e^{-t} \end{pmatrix} \end{equation} but it is not closed under matrix multiplication.
So $X=\frac{d}{dt}\gamma(t)|_{t=0}=\begin{pmatrix}1&1\\0&-1\end{pmatrix},$ but the integration of $X$ isn't equal to the curve $\gamma(t)$.