Are nilpotent Lie groups unimodular?

Question

The continuous homomorphism $\Delta:G \rightarrow \mathbb{R}^{\times}_+$ is defined by \begin{equation*} \int_G f(xy)dx = \Delta(y)\int_Gf(x)dx \end{equation*} where $dx$ is a left Haar measure on $G$. $G$ is called unimodular if and only if $\Delta(y) = 1$.

Are nilpotent Lie groups unimodular?

Answer 1

Yes, locally compact nilpotent groups are unimodular.

In fact, a locally compact group $G$ is unimodular if and only if $G/Z(G)$ is unimodular where $Z(G)$ is the center of $G$ -- see L. Nachbin, The Haar integral, corollary to Proposition 25, page 92. Abelian groups are unimodular and a simple induction on the length of the central series now shows that $n$-step nilpotent groups are unimodular.

Solvable groups are not unimodular in general, as the standard example of the $(ax + b)$-group shows.