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.

  • $\begingroup$ I have a suspicion that it should be a nested sequence of normal subgroups, if only to make sense of the LHS. $\endgroup$ – Shaun Apr 17 at 16:28
  • $\begingroup$ Its hard to answer this question because there is no context. That is: the answer is "no", and I am sure that there are many ways of seeing this. The way I have in mind is about residually finite groups. However...because the question is lacking context I do not know if this is too complicated or not! $\endgroup$ – user1729 Apr 17 at 16:29
  • $\begingroup$ (Note: f.g. nilpotent groups are residually finite, so the nilpotent case seems harder.) $\endgroup$ – user1729 Apr 17 at 16:37
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    $\begingroup$ @Shaun $HK$ is always defined as a set, you only need normality for $HK$ to be a group, which is not needed here. $\endgroup$ – Mike Earnest Apr 17 at 16:50
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    $\begingroup$ I have corrected the question. $\endgroup$ – QMath Apr 20 at 15:00

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:

  1. free groups are residually finite, and
  2. 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.


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