Let $G$ be a locally-compact topological group. Let $\Gamma$ be a discrete subgroup [Perhaps $G$ is hausdorff in order for $\Gamma$ to be closed and therefore have a haar-measure, but It was not mentioned in the origin book].
A book I'm currently reading states trivially that $\Gamma$ must be unimodular, but I fail to understand it.
It is clear to me that finite sets $E_{1,2}\subset\Gamma$ with $|E_1|=|E_2|$ must have the same measure (trivial for singletons by left-invariance of $\mu_\Gamma$, and by additivity also for finite sets). therefore $\mu_\Gamma(Eh)=\mu_\Gamma(E)$ for all finite $E\subset \Gamma$).
Now, if $\Gamma$ is finite we are done.
If it is countable, we will get by finiteness of haar measure that the measure must be the zero measure, which is a troubling result by itself.
If it is more than countable, I don't know what to do but maybe follow the same argument for countable and get the same contradiction.
p.s: I also tried using outer-regularity, but it seemed much less useful.
Best Answer
Every discrete group is unimodular. This is easy to see. If $\Gamma$ is discrete, there is a unique Haar measure $\mu$ on $\Gamma$ such that every point has measure $1$. Then $\mu_g$, the measure obtained from $\mu$ by right translation by $g$, has the same property for each $g\in G$, so that $\mu$ is also a right-invariant measure.
On the other hand, since you are referring to a locally compact topological group $G$, there is another result (perhaps you are referring to this result):
Proposition: If a locally compact group has a lattice $\Gamma$, then $G$ is unimodular.
Proof: See here, Proposition $1.3$.