I read the following theorem from J. Milne’s notes on Shimura Varieties:

Let $G$ be a reductive group over $mathbb{Q}$ and let $Gamma$ be an arithmetic subgroup of $G(mathbb{Q})$. Then

1.) The space $Gamma setminus G(mathbb{R})$ has finite volume if and only if $operatorname{Hom}(G , mathbb{G}_{m}) = 0$ (in particular, $Gamma setminus G(mathbb{R})$ has finite volume if $G$ is semisimple).

2.) the space $Gamma setminus G(mathbb{R})$ is compact if and only if $operatorname{Hom}(G , mathbb{G}_{m}) = 0$ and $G(mathbb{Q})$ contains no unipotent element other than 1.

Immediately after this theorem, there is the following example:

Let $B$ be a quaternion algebra over $mathbb{Q}$ such that $B otimes_{mathbb{Q}} mathbb{R} simeq M_{2}(mathbb{R})$, and let $G$ be the algebraic group over $mathbb{Q}$ such that $G(mathbb{Q}) = { b in B : operatorname{Nm}(b) = 1}$. The choice of an isomorphism $N otimes_{mathbb{Q}} mathbb{R} rightarrow M_{2}(mathbb{R})$ determines an isomorphism $G(mathbb{R}) rightarrow operatorname{SL}_{2}(mathbb{R})$, and hence an action of $G(mathbb{R})$ on $mathcal{H}$. Let $Gamma$ be an arithmetic subgroup of $G(mathbb{Q})$.

If $B simeq M_{2}(mathbb{Q})$, then $G$ is isomorphic to $operatorname{SL}_{2}$, which is semisimple, and so $Gamma setminus operatorname{SL}_{2}(mathbb{R})$ (hence also $Gamma setminus mathcal{H}$) has finite volume. However, $operatorname{SL}_{2}(mathbb{Q})$ contains a unipotent element $begin{pmatrix} 1 & 1 \ & 1 end{pmatrix}$, and so $Gamma setminus operatorname{SL}_{2}(mathbb{R})$ is not compact.

If $B$ is not isomorphic to $M_{2}(mathbb{Q})$, then $B$ is a division algebra and so $G(mathbb{Q})$ contains no unipotent element $neq 1$. Therefore, $Gamma setminus G(mathbb{R})$ is compact. In this way we get compact quotients $Gamma setminus cal{H}$ of $cal{H}$.

It is unclear to me how one passes between $Gamma setminus G(mathbb{R})$ and $Gamma setminus cal{H}$. Specificially, in the case when $B simeq M_{2}(mathbb{Q})$, why does $Gamma setminus operatorname{SL}_{2}(mathbb{R})$ having finite volume imply that $Gamma setminus mathcal{H}$ has finite volume? Also, I assume that (again, in the case where $B simeq M_{2}(mathbb{Q})$) that $Gamma setminus operatorname{SL}_{2}(mathbb{R})$ not being compact implies $Gamma setminus cal{H}$ is not compact. Is this true and, if so, why? In the same spirit, when $B$ is a division algebra why does $Gamma setminus G(mathbb{R})$ being compact imply that $Gamma setminus cal{H}$ is compact?