It is a known fact in the theory of non-linear symmetry realizations in physics that we can decompose a group element as (sum signs are implied)
$$ g=e^{\theta_\alpha X^\alpha}e^{u_i t^i} $$ where $t^i$ are generators of a subgroup $H \subset G$, and $X^\alpha$ are an orthonormal basis completion of G which may be thought of 'generating' the left coset $G/H$.
I tried to convince myself that this can always be done inspecting
$$ e^{\beta_\alpha X^\alpha + v_i t^i}=e^{\theta_\alpha X^\alpha}e^{u_i t^i}. $$
I proceed to use the BCH formula to obtain an equation very similar to (up to sign and other unimportant mistakes)
$$ \beta_\alpha X^\alpha + v_i t^i =\theta\cdot X + u\cdot t+ \frac{\theta^a u^b}{2} c_{abc}X^c + \frac{u^a u^b \theta^c}{12} c_{bcd} c_{ade}X^e + \frac{\theta^a \theta^b u^c}{12} c_{bcd} c_{ade}T^e +\mathcal{O}(\sqrt{u\theta}^4), $$
where $\{T\}\equiv\{X,t\}$, and I have used that the $X$ furnish a representation of $\mathfrak{h}$: $[t^a,X^b]\sim X$.
Now, the equation above is polynomial in $\theta$ and $u$, so that we should be able to find a solution over $\mathbb{C}$ for every pair of vectors $(\beta,v)$. But is it right? If this is enough, then it seems that the fact that $[t^a,X^b]\sim X$ wasn't needed.
And what about the fact that we have an infinite number of terms? Is this proposition supposed to be true only approximately (i.e., close enough to the identity)? And what if we want a solution with real $(\theta,u)$?