Nilpotent Lie group

Remarks

A connected Lie group is nilpotent if and only if its Lie algebra $\mathfrak{g}$ is a nilpotent Lie algebra, that is, the series $\mathfrak{g} \supseteq [\mathfrak{g}, \mathfrak{g}] \supseteq [\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]] \supseteq \cdots$ reaches $0$.

Example: the group $N$ of real upper triangular $3 \times 3$ matrices with $1$ on the diagonal. Writing $[a, b, c]$ for the matrix with entries $a$, $b$ above the diagonal and $c$ in the corner, a direct computation shows that the commutator of $[a, b, c]$ and $[a', b', c']$ is $[0, 0, ab' - a'b]$. So $C^2 N = \set{[0, 0, c]}$, which is the centre of $N$, and $C^3 N = \set{e}$; $N$ is nilpotent.

Non-example: the group $G$ of affine maps $x \mapsto a x + b$ of $\R$ with $a > 0$, written $(a, b)$. The commutator of $(a, 0)$ and $(1, t)$ is $(1, (a - 1)t)$, so $C^2 G$ is the group of all translations $(1, b)$, and then $C^3 G = [G, C^2 G]$ again contains every translation. So $C^k G$ is the translation group for all $k \ge 2$ and never trivial: $G$ is solvable but not nilpotent.