Intersection of a nested sequence of subgroups Let $G$ be a f.g. torsion-free nilpotent group and $(H_n)_{n\geq 1}$ a nested sequences of subgroups of finite index with trivial intersection. 

Question: Is it true that $\displaystyle\bigcap_{n\geq 1} H_nK=K$ for every subgroup $K<G$?

This question is not from homework or textbook, it is just a question from my mind, I don't know how difficult it is. I have checked several books on nilpotent groups and still was not able to get an answer. 
In general, it would be interesting to know for which classes of groups the answer is positive. 
 A: In your original question you didn't have the assumption of nilpotence. I want to answer the question both with and without this assumption, because it is easier without the nilpotency assumption an both proofs use the same idea. The answer is "no" in both cases.

A group $G$ is residually finite if for every $g\in G$ there exists a homomorphism $\phi_g: G\rightarrow F_g$ where $F_g$ is a finite group and where $\phi_g(g)\neq1$. Equivalently (by looking at the kernels of the maps $\phi_g$), the intersection of all finite index subgroups is trivial. A quick google should convince you that:


*

*free groups are residually finite, and 

*there are finitely generated, non-residually finite groups.


So let $Q$ be finitely generated and non-residually finite. There then exists a free group $\mathcal{F}$ with normal subgroup $K$ such that $\mathcal{F}/K\cong Q$. As $\mathcal{F}$ is residually finite we can take the trivially-intersecting subgroups $H_n$ to be all the subgroups of finite index. Note that every finite-index subgroup of $\mathcal{F}/K$ is the image of an $H_n$, and hence has the form $H_nK$. As $\mathcal{F}/K$ is not residually finite, there exists a non-trivial element $gK\in \mathcal{F}/K$ such that $gK\in\cap H_nK$. Hence, $\cap H_nK\neq K$ as required.
You can define yourself out of this problem: a group $G$ is called (locally) extended residually finite (LERF/ERF) if for any (finitely generated) subgroup $K<G$ and any non-trivial element $g \in G\setminus K$ there is a homomorphism $\phi_g: G\rightarrow F_g$ where $F_g$ is a finite group and where $\phi_g(g)\not\in \phi_g(K)$. What the above is saying is that free groups are not ERF. On the other hand, free groups are LERF (and LERF tends to be studied more).

So, nilpotent groups. Finitely generated nilpotent groups are residually finite, and as quotients of nilpotent groups are also nilpotent we cannot use the same argument as above. However, you can restrict to a smaller class of subgroups $H_n$:
Fix a prime $p$ and let $\mathcal{P}$ denote the class of finite $p$-groups. Suppose that $G$ is a non-cyclic, free nilpotent group. Then $G$ is residually $\mathcal{P}$ (this is due to Gruenberg, K. W. (1957), Residual Properties of Infinite Soluble Groups. Proc. Lond. Math. Soc. doi*). Hence,  the subgroups of $G$ of index some power of $p$ intersect trivially: $$\bigcap_{[G:H_n]=p^j,\; j\in\mathbb{N}}H_n=1.$$
As $G$ is free nilpotent it contains a subgroup $K$ such that $G/K$ contains a non-trivial element $gK$ of order $q$ where $q$ is coprime to $p$. Hence, $$\begin{align*}gK\in&\bigcap_{[G:H_n]=p^j}H_nK\\\Longrightarrow K\neq&\bigcap_{[G:H_n]=p^j}H_nK.\end{align*}$$
Hence, the answer is still "no" if we assume finitely generated nilpotent.

*I spend a good hour trying to first find and then access this reference. Despite the paper being >60 years old, and despite me being at my work computer, it is behind a paywall I cannot bypass; I had to use MathSciNet to work out the results. Both my current employer and my previous employer were respectable UK universities. They both have people on the LMS council. Neither university subscribes to any of the LMS journals.
