TITLE: How bad can $\pi_1$ of a linear group orbit be?
QUESTION [23 upvotes]: Let $G$ be a simply connected Lie group and $\mathcal O= G(v)=G/G_v$ a $G$-orbit in some finite-dimensional $G$-module $V$. By the homotopy exact sequence, its fundamental group $\Gamma$ is the component group of the stabilizer $G_v$:
$$
\Gamma= \pi_1(\mathcal O) = \pi_0(G_v) = G_v/G_v^{\mathrm o}.
$$

Can one characterize the class of discrete groups $\Gamma$ that occur in this way (for some orbit in some representation of some $G$)?
  Failing that, what's the most complicated $\Gamma$ you've seen occur?

My (very possibly misguided) hunch is that e.g. $\mathbf{SL}(2,\mathbf Z)$, or the discrete Heisenberg group, don't occur "because I would have heard of it". On the other hand, it is clear that:
1. Every finite group $\Gamma$ occurs.
Indeed, using e.g. its regular representation, we can embed $\Gamma$ as a closed subgroup of $G =\mathbf{SU}(N)$ for some $N$; and then the Palais-Mostow theorem guarantees [e.g. Bourbaki, Lie Groups, IX.9.2, Cor. 2] that $\Gamma$ is the stabilizer of some vector in some $G$-module, q.e.d. 
2. $\mathbf Z^n$ occurs.
Indeed, letting the additive group $G=\mathbf R^n$ act on $V=\mathbf C^n$ by
$
\smash{\begin{pmatrix}
e^{2\pi ig_1}\\
&\ddots\\
&&e^{2\pi ig_n}
\end{pmatrix}}
$
it is clear that $v={}^t(1,\dots,1)$ has stabilizer $\mathbf Z^n$.

REPLY [2 votes]: Here is a geometric proof which shows that the fundamental group of a linear group orbit must be finite by abelian (confirming the most general form of the two answers given above).
By theorem of Mostow mentioned in this question
Homogeneous manifold deformation retracts onto compact submanifold
["Covariant Fiberings of Klein spaces" Mostow 1955]
If G and G' both have finitely many connected components (for example if they are algebraic groups) then G/G' is a vector bundle over K/K' where K and K' are maximal compacts.
This is the case here because the image of a representation is always an algebraic group and the stabilizer of a vector is always Zariski closed and thus always an algebraic group.
The total space of a vector bundle is homotopy equivalent to the base. Thus G/G' is homotopy equivalent to K/K'.
Now using the fact that K has a universal cover which is a product of a compact and an abelian factor we can argue as in
https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group?noredirect=1#comment9009256_4321106
that the fundamental group of K/K' , and thus of G/G', is indeed finite by abelian
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