I'm reading Godement's paper Domaines fondamentaux des groupes arithmetiques and am confused about a part in a proof. The setting is $k$ is a number field, $\mathbb A$ is the ring of adeles of $k$, $H$ is a closed subgroup of an algebraic group $G$, both defined over $k$, and $H(\mathbb A)^0$ is the subgroup of $h \in H(\mathbb A)$ for which $|\chi(h)| = 1$ for all $k$-rational characters $\chi$ of $H$.
In the attached picture, Godement is proving that $H(\mathbb A)^0/H(k)$ is homeomorphic to its image in $G(\mathbb A)/G(k)$, and that this image is closed.
Godement uses the construction of the quotient variety $G/H$: there is a finite dimensional $k$-vector space $V$, a $k$-rational representation $\rho: G \rightarrow \operatorname{GL}(V)$, and a vector $a \in V$, for which $H$ is the stabilizer of the line through $a$.
I'm confused about the claim that $\mathbb A_k^{\ast 0} G(k).a$ closed in $V(\mathbb A)$, where $\mathbb A_k^{\ast 0}$ is the group of ideles $x$ with $|x|= 1$. Godement says that this is because $\mathbb A_k^{\ast 0}$ is the product of a compact set $Z \subset \mathbb A_k^{\ast}$ and $k^{\ast}$, and $k^{\ast} G(k).a$ is closed in $V(\mathbb A)$ because it is discrete.
My question is, how do we know that $k^{\ast}G(k).a$ is closed or discrete in $V(\mathbb A)$? Neither of these assertions seems evident to me. Even if it was discrete, that does not mean it's closed, right?
Edit: If we think of the line $\ell$ through $a$ as an element of projective space $\mathbb P(V)$, then $G/H$ is isomorphic as a variety to the image $G.\ell \subset \mathbb P(V)$, which is a locally closed subvariety of $\mathbb P(V)$.
Then I suppose $k^{\ast} G(k).a$ is the preimage of $G.\ell$ under the natural map $V(k) - \{0\} \rightarrow \mathbb P(V)(k)$. However, as we do not know that $G.\ell$ is closed in $\mathbb P(V)(k)$ (since $H$ is not necessarily parabolic), this doesn't imply that $k^{\ast} G(k).a$ is closed in $V(k)$.