# Possible set of elements generated by the Baker Campbell Hausdorff formula (BCH)

Let $$\mathfrak{g}$$ be a Lie algebra of Lie group $$G$$. Consider two sets $$S_1$$ and $$S_2$$ consisting of elements from $$\mathfrak{g}$$. Now suppose that $$[S_1, S_1]= S_1$$, which will mean that elements in set $$S_1$$ form a Lie subalgebra, while in general $$[S_2, S_2] \not\subset S_2 \text{ and }[S_2, S_2] \not \subset S_1$$ . Also suppose that $$[S_1, S_2]= \mathfrak{g}$$. In other words, elements from both sets form the entire Lie algebra.

Now, consider the following $$z=log(e^x e^y)$$ where $$x\in S_1$$ and $$y \in S_2$$. I would like to show that in general $$z\not \in S_1$$ and $$z \not \in S_2$$. It seems to me that the proof should formalize the following intuition:

1. Quite clearly, $$z$$ will in general not be an element of $$S_1$$ since BCH will involve commutators with elements from $$S_2$$.

2. Also, for any $$x \in S_1$$ and $$y \in S_2$$, $$z$$ will be in a subalgebra generated by $$x$$ and $$y$$; since the commutators between elements in $$S_1$$ and $$S_2$$ together form the entire Lie algebra there will be a some $$x \in S_1$$ and $$y \in S_2$$ such that $$z \not \in S_1$$ and $$z \not \in S_2$$.

3. If it was possible that for all $$x$$ and $$y$$ , $$z \in S_1$$ or $$z \in S_2$$, then (I think) the $$\log$$ would not be an isomorphism between sets in the neighborhood of the $$0$$ and sets in the neighborhood of $$I$$.

I want to set it up by contradiction, but I don't know how to deal with the sets rather than just the existence of a specific elements $$x$$ and $$y$$.