I was stuck on showing the following problem:
For $G$ a real connected solvable Lie group, the commutator group $[G,G]$ is nilpotent.
There are a few ways I thought about this problem:
Approach 1: By Ado's theorem, we can assume that the Lie algebra $\mathfrak{g}$ of $S$ is a matrix Lie algebra. Lie's theorem tells us there exists a complex vector space $V$ and a representation $\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ with respect to which the elements of $\mathfrak{g}$ are upper triangular. It then follows that the derived algebra $[\mathfrak{g}, \mathfrak{g}]$ is nilpotent.
The part that makes me worry is that I am assuming that $[G,G]$ has $[\mathfrak{g}, \mathfrak{g}]$ its Lie algebra.
Approach 2: Someone suggested that I can think of $G$ as a subgroup of the real linear group in which case we do get that $\mathfrak{g}$ consists of upper triangular matrices. I am unsure why we can assume that $G$ is linear
Best Answer
Ado's theorem is hard, but it's much easier to use the adjoint representation instead. The argument shows that, denoting by $\mathfrak{z}$ the center of $\mathfrak{g}$ (= kernel of adjoint representation), the linear Lie algebra $[\mathfrak{g}/\mathfrak{z},\mathfrak{g}/\mathfrak{z}]$ is nilpotent. Since $[\mathfrak{g},\mathfrak{g}]$ is a central extension of $[\mathfrak{g}/\mathfrak{z},\mathfrak{g}/\mathfrak{z}]$ (by the central kernel $\mathfrak{z}\cap[\mathfrak{g},\mathfrak{g}]$), it follows that $[\mathfrak{g},\mathfrak{g}]$ is nilpotent.
This argument also applies directly to the connected Lie group, once you know that the kernel of its adjoint representation is central.