For $G$ a real connected solvable Lie group, the commutator group $[G,G]$ is nilpotent.

I was stuck on showing the following problem:

For $$G$$ a real connected solvable Lie group, the commutator group $$[G,G]$$ is nilpotent.

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

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.
• Do you think the Adjoint representation could be used to show that $G$ is a real linear matrix group ? – user135520 Mar 25 at 13:47