Let $G\_n$ be the Lie group consisting of $n \times n$ upper triangular matrices of determinant $1$ with real entries. In other words,
$$G\_n = \{\text{$\left(\begin{matrix} a\_{11} & a\_{12} & a\_{13} & \cdots & a\_{1n} \\ 
0 & a\_{22} & a\_{23} & \cdots & a\_{2n} \\
0 & 0 & a\_{33} & \cdots & a\_{3n} \\
\vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 0 & \cdots & a\_{nn} \end{matrix}\right)$ $|$ $a\_{ij} \in \mathbb{R}$ and $a\_{11} \cdots a\_{nn} = 1$}\}.$$
Does $G\_n$ contain a lattice (i.e. a discrete subgroup of finite covolume)? The obvious thing to try is to take the subgroup consisting of matrices with integer entries, but that does not work (though it would work if we were working with strictly upper triangular matrices).



Lattice in a certain Lie group
@Edward, here is a short proof which I wrote awhile ago for my notes on group theory. 


Lemma. If a 2nd countable locally compact group $G$ contains a lattice $\Gamma$ then $G$ is unimodular. 


Proof. For arbitrary $g \in G$ consider the push-forward $\nu=R\_g(\mu)$ of the (left) Haar measure $\mu$ on $G$; here $R\_g$ is the right multiplication by $g$:
$$
\nu(E)= \mu(Eg).
$$
Then $\nu$ is also a left Haar measure on $G$. By the uniqueness of Haar measure, $\nu= c\mu$ for some constant $c > 0$. The lattice $\Gamma \le G$ has a fundamental domain $D\subset G$, i.e., a measurable subset of $G$ such that
$$
\bigcup\_{\gamma\in \Gamma} \gamma D= G, \quad \mu(\gamma D\cap D)=0, \quad \forall \gamma\in \Gamma \setminus 1.
$$
(Proof of existence of such domain uses 2nd countable assumption.) In particular, $0<\mu(D)<\infty$, since $\Gamma$ is a lattice. Then $Dg$ is again a fundamental domain for $\Gamma$ and, thus, $\mu(D)=\mu(Dg)$. Hence, $\mu(D) = \mu(Dg) = c\nu(D)$. It follows that $c=1$. Thus, $\mu$ is also a right Haar measure. qed


Another proof could be found in Chapter I of Raghunathan's book "Discrete subgroups of Lie groups". 


