Consider a locally compact group $G$ and a closed subgroup $H$ of $G$, and let $G$ act on $G/H$ by left translation. Green's imprimitivity theorem implies that the crossed product $C_0(G/H)\rtimes G$ is Morita equivalent to $C^*(H)$. With more work, one shows that the crossed product is isomorphic to $C^*(H)\otimes K(L^2(G/H))$.
Is there a simpler direct proof of the above for the reduced crossed product in the case where $G$ is discrete? I can do the case of $G$ acting on itself by left translation, but can I transfer the idea to the case above?
Maybe computing an explicit example may help me. So I considered $\mathbb{Z}_4$ acting on $\mathbb{Z}_4/\mathbb{Z}_2$, and the above tells me that I should be getting $M_2(\mathbb{C})\otimes \mathbb{C}^2$. My intuition tells me that this is because 2 acts the same way as 0, while 3 acts the same way as 1. I've been trying to write down an explicit isomorphism but I haven't been able to do so.
Also, suppose $K$ is a finite subgroup of a discrete group $G$, and $S$ is a $K$-space. Let $K$ act on $G\times S$ by $k(g,s)=(gk^{-1},ks)$, and let $G\times_KS$ be the orbit space. $G$ acts on $G\times_KS$ by $g[h,s]=[gh,s]$. When $G$ acts properly on some locally compact Hausdorff space $X$, every point in $X$ has a $G$-invariant open neighborhood that is $G$-homeomorphic to one of these ''twisted products''.
How do we describe the (reduced) crossed product of $G\times_KS$ by $G$ up to isomorphism or Morita equivalence? It seems to me that the action of $G$ on $G\times_KS$ looks very much like the action of $G$ on $G/K$ so it should be somehow related to the result above.