# Subgroups of residually finite groups

Let $\Gamma$ be a finitely generated residually finite group. For a subgroup of finite index $\Lambda<\Gamma$ let us denote by $\pi_\Lambda:\Gamma\rightarrow \Gamma/\Lambda$ the quotient map. Is it possible to find a subgroup $H<\Gamma$ such that the restriction of $\pi_\Lambda$ to $H$ is surjective for every finite index subgroup $\Lambda<\Gamma$?

Does $SL_n(\mathbf Z)$ have such subgroups?

Can we find such $H$ finitely generated?

• You're asking for subgroups which are dense in the profinite topology. There are many constructions of groups $\Gamma$ with such subgroups, even finitely generated. By the congruence subgroup property and strong approximation, $SL_n$ has such fg subgroups when $n$ is at least 3. There are no such fg subgroups when $n=2$, since $SL_2$ is virtually free, hence LERF. – HJRW Jan 6 '17 at 16:24
• To complete Henry's answer, $\Gamma=SL_2(\mathbf{Z})$ has such non-fg subgroups (e.g., the kernel of any quotient map from $\Gamma$ to an infinite simple group). – YCor Jan 6 '17 at 17:56
• For a longer discussion and obstructions (other than LERF) of the existence of such f.g. $H$, see Section 3 of my survey arxiv.org/abs/math/0509090. – YCor Jan 6 '17 at 17:58

A general class of such groups are maximal subgroups $H<\Gamma$ of infinite index. Such a subgroup $H$ must surject any finite quotient of $\Gamma$. For finitely generated linear groups like $SL_n(\mathbb{Z})$ which are not virtually solvable, maximal subgroups of infinite index were constructed by Margulis-Soifer. I don't know in what generality it is known which residually finite groups have maximal subgroups of infinite index (and this is a much stronger property than being dense in the profinite topology). Note, however, that if $H<\Gamma$ is dense in the profinite topology, then a maximal proper subgroup $H < K < \Gamma$ containing $H$ must be of infinite index (since $H$ is not contained in any finite-index proper subgroup, and hence $K$ cannot be finite-index). So the question boils down to which finitely generated residually finite groups contain a maximal subgroup of infinite index (however, note that the existence of maximal subgroups in a finitely generated group requires Zorn's lemma see the comments for how to find a maximal subgroup containing $H$).
If $\Gamma$ is LERF (like a finitely-generated Kleinian group), then $H$ must be infinitely generated to have this property. However, for $n>2$, as Henry indicates in his comment, one has many finitely-generated subgroups of $SL_n(\mathbb{Z})$ which are dense in the profinite topology. Finitely generated subgroups are contained in a proper finite-index subgroup is called the engulfing property. Your question is about whether $\Gamma$ has the engulfing property for all subgroups $H$.
Remark: Incidentally, in your statement, you don't assume that $\Lambda$ is normal in $\Gamma$. But since $\Gamma$ is finitely generated, there is an normal subgroup $Core(\Lambda)= \cap_{g\in\Gamma} g\Lambda g^{-1}$ which is also of finite index, so that there is a factorization $\Gamma \to \Gamma/Core(\Lambda)\to \Gamma/\Lambda$. Then if $H$ surjects $\Gamma/\Lambda$ for every normal $\Lambda\lhd \Gamma$ of finite index, then $H$ surjects $\Gamma/\Lambda$ for every finite-index subgroup $\Lambda < \Gamma$. I was using this implicitly in the first paragraph.
• No, you enumerate elements $(g_n)$ of some free group mapping onto $G$, and define $G_0=\{1\}$ and $H_{n}=\langle G_n,g_{n+1}\rangle$, and $G_{n+1}$ to be either $G_n$ or $H_n$ according to whether $H_n=G$. – YCor Jan 6 '17 at 23:02
• @YCor Okay, then the maximal subgroup is $\cup_n G_n$, and I guess it's clear that this is maximal and (using finite generation) proper. I guess it's also clear that you can find a maximal subgroup containing any subgroup by taking $G_0=H$ instead. Moreover, you can enumerate the elements of $G$, you don't need to enumerate the elements of a free group mapping onto $G$ (at least, I don't see the need for doing this). – Ian Agol Jan 7 '17 at 0:10
• To enumerate elements of $G$ without countable choice requires another little argument; I was just avoiding it. – YCor Jan 7 '17 at 9:10