Simple module over $ֿ\mathbb{Z}G$ has a $\mathbb{Z}N$ composition series when $N \triangleleft G$ is nilpotent and of finite-index Assume that $M$ is a simple $\mathbb{Z}G$ module with $G$ finitely-generated, virtually nilpotent group.
Take $N\leq G$ to be a normal, finite index, nilpotent subgroup.
The claim is that $M$ has a finite-length composition series as a $\mathbb{Z}N$-module. How can I see this?
One approach I've tried: I know that a module has a finite-length composition series if and only if the module is both Artinian and Noetherian. Now in fact I can prove (with some "heavier" theorems) that $M$ is Noetherian in this case (this follows because $N$ is also finitely generated, and nilpotent, hence polycyclic, in essence). But, this approach doesn't use the fact the $M$ is simple as a $\mathbb{Z}G$-module and moreoever I can't prove the descending-chain-condition anyways.
 A: If $R$ is a ring, $G$ a group, and $N$ a normal subgroup of finite index, then any simple $RG$-module is semisimple and of finite length (at most $[G:N]$) as an $RN$-module.  No other hypotheses about $G,N$ are needed.  This result can be found in Passman's "The Algebraic Structure of Group Rings", Theorem 7.2.16.  He cites a 1937 paper of A. H. Clifford.
Even more generally, if $A,B$ are rings such that $B$ is a finite normalizing extension of $A$, then any simple $B$-module is semisimple and of finite length as an $A$-module.  This is Proposition 10.1.9(ii) in "Noncommutative Noetherian Rings" by McConnell and Robson -- see paragraph 10.1.5 for notation and assumptions.  To say $B$ is a finite normalizing extension of $A$ means that there are elements $b_1,\dots,b_k$ that generate $B$ as a left (or right) $A$-module and satisfy $Ab_i=b_iA$ for each $i$.
Here is a sketch of the proof in the group ring case.  Let $g_1,\dots,g_k$ be a complete set of coset representatives for $N$ and let $M$ be a simple left $RG$-module.  It's easy to see $M$ is finitely generated as an $RN$-module, so it contains a maximal $RN$-submodule $L$.  The intersection $\cap_{i=1}^k g_iL$ is an $RG$-submodule of $M$ and so must be zero.  This implies that $M$ embeds as an $RN$-submodule in the direct sum of the $RN$-modules $M/g_iL$.  Since $g_iL$ is a maximal $RN$-submodule of $M$, each of these factor modules is simple.  Hence $M$ is semisimple as an $RN$-module, of length at most $k$.
