Prove the flow of the left-invariant vector field $w_X$ at time $t$ is$\Phi_X^t(g)=g\exp (tX), t \in \mathbb R, g \in G$

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If $M$ is a smooth manifold, $G$ a Lie group and $v, w$ are vector fields. For small $t \in \mathbb R$ there is a flow determined by $v$, $\phi^t:M \rightarrow M$ which is a diffeomorphism. The pullback of $w$ along $(\phi^t)^*$ is $$(\phi^t)^*w(p)=(T_p\phi^t)^{-1}w(\phi^t(p)), \forall p \in M$$ and the vector field

$[v,w]:= \frac{d}{dt}(\phi^t)^*w|_{t=0}$

If $X \in T_e(G)$, we have the left-invariant vector field $w_X(g):=T_e(l_g)X$, where $l_g$ is the map left-multiplication by $g$.

I am trying to write the flow of the left-invariant vector field $w_X$ at time $t$, denoted $\Phi_X^t$ in terms of the exponential map:

$\Phi_X^t(g)=g\exp (tX), t \in \mathbb R, g \in G$

but I have no idea how to get started,since differential geometry was mainly a self-taught subject, I am not very familiar with manipulations, for instance I don't know how to deal with the $(T_p\phi^t)^{-1}$, although I know how the diffeomorphism is defined.

Could you show me how should I proceed as an example?

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Fix $g \in G$. Differentiate the curve $\gamma(t) := \Phi_X^t(g) = g\exp(tX), \ t \in \mathbb{R}$: $$\frac{d}{dt}|_{t=t_0} \gamma(t) = \frac{d}{dt}|_{t = 0}\gamma(t+t_0) = \frac{d}{dt}|_{t = 0}(g\exp(t_0X+tX)) = \frac{d}{dt}|_{t = 0}(g\exp(t_0X)\exp(tX)) = \frac{d}{dt}|_{t = 0}(\gamma(t_0)\exp(tX)) = T_e(l_{\gamma(t_0)})(X) = w_X(\gamma(t_0))$$ Therefore, $\gamma$ is the integral curve of the vectorfield $w_X$. By uniqueness the claim follows.