I want to prove the following proposition about lie groups and Lie algebras:
Let $G$ be a connected Lie group, and $\mathfrak{g}$ be the Lie algebra of G. Then, the Lie algebra $\mathfrak{g}$ is solvable if and only if $G$ has a subnormal series whose factor groups are all Abelian, with any term being connected lie subgroup of G (i.e. $G$ has a sequence $\{ G_i \}$ such that $G_i$ is a connected Lie subgroup of $G$ and is a normal subgroup of $G_{i-1}$, and the quotient $G_{i-1} / G_i$ is abelian), where the definition of "$\mathfrak{g}$ is solvable" is the one using a series of ideals $\{ \mathfrak{g}_i \}$.
But I can't prove the part that $\mathfrak{g}_{i-1} / \mathfrak{g}_i$ is Abelian iff $G_{i-1} / G_i$ is Abelian. (Remark: $G_i$ is generated by the image of $\mathfrak{g}_i$ under the exponential map.)
How do I prove this? I want to know the way to prove directly, not to prove $\mathfrak{g}_{i-1} / \mathfrak{g}_i$ is the Lie algebra of $G_{i-1} / G_i$.