I am a math student who is introducing to Lie Groups. My teacher uses this result in an exercise, but he did not proved and I dont know how can I do. I would like to know how to prove it so, any help is welcome. I am sure the result is easy for some who is familiarized with Lie Group. The result is this:
Given $G$ a Lie group, $X\in \mathfrak{g}$ an element of its Lie algebra and a smooth function $f\in C^\infty (G)$, then $$ X(f)(\sigma) = \frac{d}{dt}\vert_{t=0} \ [f(\sigma \exp (tX))],\quad \forall \sigma \in G,$$ where $exp$ stands for the exponential map.
Thanks in advance.
This probably arose when explaining the equivalence between $T_e G$ and the space of left-invariant vector fields on $G$.
Let us think of $Lie(G)$ as the tangent space $T_eG$ of $G$ at the identity $e$. Note that your equation is equivalent to (I prefer the notation $V_p$ for the vector field $V$ evaluated at the point $p$) \begin{equation}X_\sigma= d/dt|_0\ \sigma \exp(tX) =d/dt|_0\ L_\sigma( \exp(tX)), \quad (*)\end{equation} with $L_\sigma$ left multiplication by $\sigma$. Since $\exp(tX)$ is a curve in $G$ starting at $e$ with tangent $X$, (*) can be rewritten \begin{equation} X_\sigma = L_{\sigma *} X.\qquad (**) \end{equation}
Equation (**) associates a vector field on $G$, which is easily checked to be left-invariant, to an element of $T_eG$ (abusing notation both are denoted by $X$). Conversely, a left-invariant vector field gives an element of $T_eG$ by evaluating it at $e$. Since a left-invariant vector field is determined by its value at $e$ (or any other point, but $e$ is convenient), we obtain a bijection \begin{equation} \{\text{Left-invariant vector fields on } G\} \leftrightarrow \{\text{Elements of } T_eG \} \qquad (***).\end{equation}
This identification allows us to introduce the bracket operation on $T_eG$ and so give it the structure of a Lie algebra. In fact suppose $X_e, Y_e\in T_eG $ and denote by $X,Y$ the associated left-invariant vector fields. One can show that $Z:=[X,Y]$ is a vector field, acting on a function $f$ as $Z(f)=X(Y(f))-Y(X(f))$ (this is true on any manifold), and that $Z$ is left-invariant. By virtue of (***) one can therefore define $[X_e,Y_e]=Z_e$.
If $G$ is a matrix group there is a more concrete description of the bracket operation. One can identify $T_eG$ which a certain space of matrices (e.g. if $G=GL(n,\mathbb{R})$ then $T_eG$ is the space of $n\times n$ matrices with real entries), and it is possible to show that the bracket operation defined in terms of left-invariant vector fields is equivalent to the commutator of matrices in $T_eG$. However for a general Lie group we cannot use matrices and we need the definition via the bracket of left-invariant vector fields.
EDIT TO ANSWER THE COMMENT In the question you say that $X$ is in the Lie algebra, but I am not sure which definition of the Lie algebra are you thinking of. In my answer I assumed that the $X$ on the rhs of your equation is an element of $T_e G$. If so equation (**) should be viewed as a definition of the associated left-invariant vector field on $G$, and not something to be proved.
The only case which comes to my mind in which there is something to be proved is if you are defining $X$ to be a left-invariant vector field on $G$, with value at the identity confusingly still indicated by $X$. You can then ask what is its value at other points of $G$. Because of left invariance, the answer is then given by (**), which this times is a result rather than a definition.