# If $X,Y$ generates $\mathfrak{g}$ then $e^{tX}$ and $e^{tY}$ generates de Lie Group G.

I'm trying to solve the following problem of the book "Grupos de Lie - Luiz A. B. San Martin":

Question: Let $$G$$ be a connected Lie Group with Lie algebra $$\mathfrak{g}$$. Suppose that $$X,Y \in \mathfrak{g}$$ generate $$\mathfrak{g}$$ (i.e. $$X$$ and $$Y$$ are not contained in any proper subalgebra of $$\mathfrak{g}$$, or in a equivalent way, the successive brackets of $$X$$ and $$Y$$ generates the Lie Algebra $$\mathfrak{g}$$). Show that the $$1$$-parameter groups $$\text{exp}(tX)$$ and $$\text{exp}(tY)$$ generate $$G$$.

Consider $$B = \langle \exp (t X), \exp (s Y)\rangle$$ as the smallest group $$\subset$$ $$G$$, such that $$\exp (tX)$$ and $$\exp (sY)$$ belongs to $$B$$ , $$\forall\ t,s \in \mathbb{R}$$.

## Some ideas.

Once $$G$$ is connected the problem is equivalent to showing that in a neighborhood $$V$$ of $$1$$ in $$G$$, the result holds. So I'm trying to find a smart way to generate the brackets of $$X$$ and $$Y$$, by derivating curves contained in $$\langle \exp (tX), \exp (s Y)\rangle$$ in $$t=0$$. For example, something like the curve $$\exp(X) \exp(tY) \exp(-X)$$, in this specific case

\begin{align*}\left.\frac{d}{dt}\exp(X) \exp(tY) \exp(-X) \right|_{t=0} &=\left.\exp(te^{ad(X)}Y) \right|_{t=0}\\ &= e^{ad(X)}Y\\ &= Y + [X,Y] + \frac{1}{2}[X,[X,Y]] + ..., \end{align*}

If I would be able to construct curves such that $$\alpha_i: \mathbb{R} \to G$$, such that $$\text{Im}(\alpha_i)\subset \langle \exp (tX), \exp (sY)\rangle,$$ $$\alpha_i(0) = 1$$ and $$\left\{\left.\frac{d}{dt} \alpha_1 (t) \right|_{t=0},..., \left.\frac{d}{dt} \alpha_n (t)\right|_{t=0} \right\},$$

is a basis of $$\mathfrak{g}$$, then the local diffeomorphism at $$0$$, \begin{align*} \varphi: \mathbb{R}^n &\to G\\ (t_1,...,t_n)&\mapsto \alpha_1(t_1)\cdot ...\cdot \alpha_n(t_n), \end{align*}

would solve the problem, however, this idea has not been fruitful.

Can anyone help me?