Question.
Let $G$ be a real Lie group of dimension $n$ and let $\mathfrak g$ denote the associated Lie algebra. Assume that the exponential map $\exp\colon \mathfrak g \to G$ is surjective. Let $\{X_1\ldots X_n\}$ be a basis of $\mathfrak g$.
Is it true that for every $g\in G$ one has $$\tag{1}g=\exp(\theta_1 X_1)\exp(\theta_2X_2)\ldots \exp(\theta_n X_n) $$ for some $(\theta_1\ldots \theta_n)\in\mathbb R^n$?
If not, can you provide a counterexample? Are there some assumptions on $\{X_1\ldots X_n\}$ that make it true?
Motivation.
Surjectivity of $\exp$ means that, for any $g\in G$, one has $(\phi_1\ldots \phi_n)\in\mathbb R^n$ such that: $$\tag{2} g=\exp(\phi_1X_1+\phi_2X_2+\ldots+\phi_nX_n).$$ If I understand things correctly, the Baker-Campbell-Hausdorff formula applied to (2) gives the existence of some coefficients $(\theta_1\ldots \theta_n)$ such that $$ \tag{3} g=\exp(\theta_{i_1}X_{i_1})\exp(\theta_{i_2}X_{i_2})\ldots \exp(\theta_{i_n}X_{i_n}).$$ The difference between (1) and (3) is that in (1) we require that the exponentials are in the prescribed order.