Let $G$ be a locally compact Hausdorff group with left modular character $\delta$. A discrete subgroup $\Gamma$ of $G$ is called a lattice if $\delta(x) = 1$ for all $x \in \Gamma$, and if the resulting left invariant measure on $G/\Gamma$ is finite.
If $G$ has a lattice, and I'm trying to understand why $G$ has to be unimodular. The proof comes down to showing that since $G/\Gamma$ has finite measure, so does $G/N$, where $N$ is the kernel of $\delta$. Here we have $\Gamma \subseteq N \subseteq G$, so this should have something to do with $G/N$ being a continuous image of $G/\Gamma$. Somehow we need to relate the measures on $G/N$ and on $G/\Gamma$.
Once we show that $G/N$ has finite measure, we use the result that if a topological group has finite Haar measure, then it is compact. Then $G/N$ surjects onto a compact subgroup of $(0,\infty)$, which must be trivial.