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?
 A: 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.
