TITLE: a categorical Nakayama lemma?
QUESTION [51 upvotes]: There are the following Nakayama style lemmata:

(the classical Nakayama lemma) Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$-module. If $m_1, \ldots, m_n$ generate $M$ modulo $I$, where $I \subset \mathrm{Jac}(R)$, then they generate $M$.
(the graded Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.
(the filtered Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.
(the topological Nakayama lemma, see [Neukirch, Schmidt, Wingberg], Cohomology of Number Fields, (5.2.18)): Let $\mathcal{O}$ be a commutative local ring complete in the $\mathfrak{m}$-adic topology with finite residue field of characteristic $p$.  Assume $G$ is a pro-$p$-group and $M$ is a compact $\mathcal{O}[[G]]$-module. If $M/\mathfrak{m}$ is a finitely generated $\mathcal{O}[[G]]$-module, so is $M$.
(Burnside's basis theorem, see also http://groupprops.subwiki.org/wiki/Burnside%27s_basis_theorem) Let $G$ be a group such that its Frattini subgroup $\Phi(G)$ is finitely generated. Then a subset of $G$ generates $G$ iff it generates it modulo $\Phi(G)$.

[tbc]
Now my question is: Is there a common categorical version, like there is a categorical generalisation of Baer's criterion (In a suitable abelian catgory, an object $I$ is injective iff for all subobjects $U$ of a generator $G$ and morphisms $U \to I$ there is a lift $G \to I$)?

REPLY [7 votes]: Let me build a little on Gjergji's answer, using his terminology, in order to get closer to how I have usually heard Nakayama's lemma stated. Suppose, in addition, that $\mathbb{Fr}_p(G)$ is finitely generated and let $S \subseteq G$. Then, if the image of $S$ $P$-generates $G/\mathbb{Fr}_p(G)$, then $S$ $P$-generates $G$. (Hmm, it's not clear to me that $\mathbb{Fr}_p(G)$ is always normal in $\mathbb{Fr}_p(G)$, although it is in all of the examples so far. So we might need to add that as a hypothesis.)
Proof: Let $x_1$, $x_2$, ..., $x_r$ $P$-generate $\mathbb{Fr}_p(G)$. Then $G=\langle S, x_1, x_2, \ldots, x_r \rangle_P$ (exercise!). Repeatedly using the definition of a nongenerator, $G = \langle S \rangle_P$. $\square$