Let $G$ be a locally-compact topological group, and $H$ be a normal subgroup.
$G/H$ is a locally-compact topological group as well, and if we assume $H$ to be closed then $G/H$ is Hausdorff and therefore admits a Haar measure $\mu_{G/H}$.
It is also known that if $\Delta_{G}|_{H}=\Delta_{H}$ (the modular functions), then $G/H$ admits a (unique) finite $G$-invariant Borel measure $\nu_{G/H}$.
What is the relation between $\mu_{G/H}$ and $\nu_{G/H}$? Under what conditions are they the same measure? Is $\nu_{G/H}$ also finite iff $G/H$ is compact?
Another related question, which also rises from me reading about Haar measures from a book about ergodic theory:
in the same context, if $\Gamma$ is a lattice (i.e discrete subgroup for which $G/\Gamma$ admits a finite $G$-invariant Borel measure) and $H$ is again a closed normal subgroup with the additional condition $\Gamma\subset H$, how can we deduce that the Haar measure $\mu_{G/H}$ is finite?
[I've given a lot of thought to this but could not get it right. I wanted to show that $G/H$ is compact, but it might require me to know that $G/\Gamma$ which might not be true since only the $G$-invariant measure is finite, and not neccessarily the Haar measure.]
P.S: Please don't use, without reference to the proof, any equivalent definitions to the object I mentioned. I'm new to the field, and would like for starters to understand everything under the formalism I've given here.
Since $G, H,$ and $G/H$ are locally compact Hausdorff groups, they all have left Haar measures $\mu_G, \mu_H, \mu_{G/H}$.
Temporarily dropping the assumption that $H$ is normal in $G$, let $G/H$ be the set of left cosets of $H$ in $G$ with the quotient topology. Then $G$ acts on $G/H$ continuously by $(g, xH) \mapsto gxH$. Let $\lambda$ be a Radon measure on $G/H$. We say that $\lambda$ is $G$-invariant if $\lambda(g \cdot E) = \lambda(E)$ for any Borel set $E$ of $G/H$ and any $g \in G$. As you mentioned, there exists a $G$-invariant Radon measure on $G/H$ if and only only if the restriction of the modular character $\delta_G$ of $G$ to $H$ coincides with the modular character $\delta_H$ of $H$. If this is the case, then such measures on $G/H$ are unique up to scalar.
Now returning to the assumption that $H$ is normal, it is easy to see that the left Haar measure $\mu_{G/H}$ is a $G$-invariant Radon measure: for $E \subseteq G/H$ Borel and $g \in G$, we have
$$\mu_{G/H}(g \cdot E) = \mu_{G/H}((gH)E) = \mu_{G/H}(E)$$
Conclusion: if $H$ is a normal closed subgroup of $G$, then the restriction of $\delta_G$ to $H$ is $\delta_H$, and a left Haar measure on $G/H$ is the same thing as a $G$-invariant Radon measure.