\section{Introduction}
\label{sect:intro}

\subsection{$N$-series} 
\label{intro:n-series}
The main goal of this paper is to analyze the behavior of lower central 
series and associated graded Lie algebras under split extensions 
of groups. We do this for both the usual lower central series, and 
its rational and mod-$p$ analogues, working for the most part 
in the unified context of $N$-series. 

The study of the lower central series and the associated 
graded Lie algebra of a group goes back to the work 
of Hall \cite{Hall} and Magnus  \cite{Magnus35, Magnus40} 
from the 1930s. In his thesis, \cite{Lazard}, Lazard greatly 
developed these concepts, and introduced the notion of 
an $N$-series, which serves as a powerful abstraction of the 
lower central series. An $N$-series for a group $G$ is a 
descending filtration by 
subgroups $\{K_n\}_{n\ge 1}$, starting at $K_1=G$ and such 
that $[K_m, K_n] \subseteq K_{m+n}$, for all $m,n\ge 1$. 
Clearly, this is a central series (and thus, a normal series), 
and so the quotient groups, $K_n/K_{n+1}$, are abelian. 
The direct sum of these quotients, $\gr^{K} (G)  
= \bigoplus_{n\ge 1} K_n/K_{n+1}$, acquires the 
structure of a graded Lie algebra, with Lie bracket 
induced from the group commutator. 

\subsection{Lower central series} 
\label{intro:lcs-series}
The prototypical example of an $N$-series is the lower central series, 
$\gamma(G)=\{\gamma_n(G)\}_{n\ge 1}$, that starts at 
$\gamma_1 (G)=G$ and is defined recursively by  
$\gamma_{n+1}(G) =[G,\gamma_n (G)]$. 
In a seminal paper from 1965, \cite{St}, Stallings established 
an important relationship between the low-dimensional 
homology of a group and its lower central series (LCS). In the 
process, he also considered the rational and mod-$p$ versions 
of this series, which we denote by $\gamma^{\rat}(G)$
and $\gamma^p(G)$, respectively. Both these series 
(which were later investigated in depth by Cochran and 
Harvey \cite{CH-gt08, CH-jlms08}), start 
at $G$ and are defined recursively by 
\begin{equation}
\label{intro:gamma}
\gamma^{\rat}_{n+1}(G) =\!\sqrt{[ G,\gamma^{\rat}_n(G)]} 
\quad\text{and}\quad
\gamma^{p}_{n+1}(G)=
\big(\gamma^{p}_{n}(G)\big)^p  \big[G,\gamma^{p}_{n}(G)\big] \, .
\end{equation}

Here, for a subset $S\subseteq G$, we let $\!\sqrt{S}$ 
denote the set of elements of $G$ which have a positive 
power that belongs to $S$, and we let $S^p$ denote 
the subgroup generated by all the $p$-th powers of 
elements in $S$. It is known that the rational and 
mod-$p$ lower central series are also $N$-series; 
moreover, the graded quotients of $\gamma^{\rat}(G)$ 
are torsion-free abelian groups, while those of $\gamma^{p}(G)$ 
are elementary abelian $p$-groups; see \cite{Passman, BL, Mass, 
Paris}. It is also known that $\gamma^{\rat}_n(G)$ coincides with the $n$-th 
dimension subgroup of $G$, defined in terms of the $I$-adic filtration on 
the group algebra $\mathbb{Q}[G]$; see \cite{Passi79, Passman, Mass}. 
The graded Lie algebras associated to 
$\gamma(G)$, $\gamma^{\rat}(G)$, and $\gamma^p(G)$ 
are denoted by $\gr(G)$, $\gr^{\rat}(G)$, and $\gr^p(G)$, 
respectively.

\subsection{Split extensions} 
\label{intro:split}
For most of the paper we focus our attention on groups 
that arise as split extensions of the form 
$B=A\rtimes_{\varphi} C$, where $A$ is a normal 
subgroup of $B$ with quotient group $C$. The projection
map $B\to C$ admits a splitting $\sigma\colon C\to B$, 
which determines (and is determined by) the monodromy 
of the extension, $\varphi\colon C\to \Aut(A)$; in particular, 
we may view $C$ as a subgroup of $B$.

In order to describe the lower central series $\gamma(B)$ and the 
associated graded Lie algebra $\gr(B)$, we consider a series 
$L=\{L_n\}_{n\ge 1}$ of normal subgroups of $A$, 
which is defined inductively by setting $L_1=A$ and letting 
$L_{n+1}$ be the subgroup generated by the 
commutators $[A,L_{n}]$, $[A, \gamma_{n}(C)]$,  
and $[L_{n},C]$.  This series was recently 
introduced by Guaschi and Pereiro, who showed 
in \cite[Theorem 1.1]{GP} that the monodromy 
of the extension restricts to maps 
$\varphi\colon \gamma_n(C)\to \Aut(L_n)$, and that 
\begin{equation}
\label{intro:gamma-b}
\gamma_n(B)=L_n\rtimes_{\varphi} \gamma_n(C)
\end{equation}
for all $n\ge 1$. We prove in Theorem \ref{thm:gp-nseries} 
that the Guaschi--Pereiro series is, in fact, an $N$-series. This additional 
information allows us to give in Theorem \ref{thm:gu-pe} 
a similar, yet more streamlined and transparent proof of their result. 
Furthermore, we show in Theorem \ref{thm:gp-gr} that the associated 
graded Lie algebra of $B$ splits as a semidirect product of graded Lie algebras, 
\begin{equation}
\label{intro:gr-b}
\gr(B) = \gr^L(A) \rtimes_{\bar\varphi} \gr(C),
\end{equation}
with monodromy $\bar\varphi\colon \gr(C) \to \Der(\gr^L(A))$  
induced by $\varphi$.  As we show by means of several examples, 
it is not possible, in general, to replace $\gr^{L}(A)$ by $\gr(A)$ 
in \eqref{intro:gr-b}.  

Nevertheless, there is an important class of split extensions 
where this is possible. Suppose $B=A\rtimes_{\varphi} C$ is an 
almost direct product of groups; that is, $C$ acts trivially on the 
abelianization $A_{\ab}=\gr_1(A)$. In this case, 
we show in Theorem \ref{thm:fr-lem} that $L_n=\gamma_n(A)$ 
for all $n\ge 1$.  When combined with formulas \eqref{intro:gamma-b} 
and \eqref{intro:gr-b}, this theorem recovers two well-known results of Falk 
and Randell  from \cite{FR}, namely, the semidirect product 
decompositions,  $\gamma_n(B)=
\gamma_n(A) \rtimes_{\varphi} \gamma_n(C)$ 
and $\gr(B)=\gr(A) \rtimes_{\bar\varphi} \gr(C)$.

\subsection{Rational LCS} 
\label{intro:rat}

Building on the approach outlined above, we establish analogous 
results for the rational and mod-$p$ lower central series. In the first 
instance, we start by showing in Proposition \ref{prop:sq-bar} that 
$\gamma^{\rat}_n(G)=\!\sqrt{\gamma_n(G)}$, for all groups $G$.  
Work of Massuyeau \cite{Mass} now implies that 
$\gamma^{\rat}(B)$ and $\!\sqrt{L}$ are 
$N$-series for $B$ and $A$, respectively.  We then show in 
Theorem \ref{thm:gp-rat} that 
\begin{equation}
\label{intro:gamma-q}
\gamma^{\rat}_n(B)=\!\sqrt{L_n}\rtimes_{\varphi} \gamma^{\rat}_n(C)
\end{equation}
for all $n\ge 1$, from which we deduce that 
$\gr^{\rat}(B)=\gr^{\!\sqrt{L}}(A) \rtimes_{\bar\varphi} \gr^{\rat}(C)$. 
The arguments we give parallel those in the integral case, though they do 
require an additional careful analysis (based on lemmas from 
\S\ref{subsec:comms} and \S\ref{subsec:comm-semi}) of the 
way commutators and powers interact in split extensions.

Noteworthy is the case of rational almost direct products, i.e., split 
extensions of the form $B=A\rtimes_{\varphi} C$, with $C$ acting trivially 
on the torsion-free abelianization $A_{\abf}=\gr^{\rat}_1(A)$. 
For such extensions, we show in Theorem \ref{theorem:fr-rational} 
that $\!\sqrt{L_n}=\gamma^{\rat}_n(A)$ for all $n\ge 1$. Using formula 
\eqref{intro:gamma-q}, we obtain rational analogues of the results 
of Falk and Randell; namely, 
\begin{equation}
\label{intro:fr-q}
\gamma^{\rat}_n(B)=
\gamma^{\rat}_n(A)\rtimes_{\varphi} \gamma^{\rat}_n(C)
\end{equation}
for all $n\ge 1$, from which we conclude in Theorem \ref {thm:q-fr}  that 
the rational associated graded Lie algebra of $B$ decomposes as 
$\gr^{\rat}(B)=\gr^{\rat} (A) \rtimes_{\bar\varphi} \gr^{\rat}(C)$.

\subsection{Mod-$p$ LCS} 
\label{intro:mod-p}
Turning now to the mod-$p$ lower central series, we define by analogy with 
the integral and rational cases a sequence $L^p$ of subgroups of $A$, 
by setting $L^{p}_1=A$ and letting $L^{p}_{n+1}$ be the subgroup 
generated by $\big(L^{p}_{n}\big)^p$, $[A,L^{p}_{n}]$,  
$[A, \gamma^{p}_n(C)]$, and $[L^{p}_{n},C]$. 
We then prove in Theorem \ref{thm:p-n-seq} that the sequence 
$L^p=\{L^p_n\}_{n\ge 1}$ forms a $p$-torsion $N$-series for $A$.  
Furthermore, we show in Theorem \ref{thm:gp-p} that 
\begin{equation}
\label{intro:gamma-p}
\gamma^{p}_n(B)=L^p_n\rtimes_{\varphi} \gamma^{p}_n(C)
\end{equation}
for all $n\ge 1$. As a consequence, we show that $\gr^{p}(B)$ decomposes 
as a semidirect product of graded Lie algebras,  
$\gr^{p}(B)=\gr^{L^p}(A) \rtimes_{\bar\varphi} \gr^{p}(C)$. 

Finally, suppose that $C$ acts trivially on $A_{\ab}\otimes \mathbb{Z}_p=\gr_1^p(A)$.  
We then show in Theorem \ref{thm:fr-lem-p}
that $L_n^p=\gamma^{p}_n(A)$ for all $n\ge 1$. Using formula 
\eqref{intro:gamma-p}, this allows us to recover in Theorem \ref {thm:fr-p} 
a result first proved by Bellingeri and Gervais in \cite{BG}, to wit,
\begin{equation}
\label{intro:fr-p}
\gamma^{p}_n(B)=\gamma^{p}_n(A)\rtimes_{\varphi} \gamma^{p}_n(C), 
\end{equation}
for all $n\ge 1$, from which we conclude that 
$\gr^{p}(B)=\gr^{p} (A) \rtimes_{\bar\varphi} \gr^{p}(C)$.

\subsection{Applications and further directions} 
\label{intro:further}

One of the many reasons for studying the lower central series and 
its rational and modular variants comes from the fact that these series 
control the corresponding residual properties of a group $G$. 
Namely, $G$ is residually nilpotent, respectively, residually torsion-free 
nilpotent, or residually $p$ if and only if the intersection of the terms 
of the series $\gamma(G)$, respectively $\gamma^{\rat}(G)$, or 
$\gamma^{p}(G)$ is trivial. 

Suppose now that $B=A\rtimes C$ is a split extension. As noted in a 
special case by Falk and Randell \cite{FR88}, if the factors $A$ and $C$ 
are residually nilpotent groups, and $C$ acts trivially on $A_{\ab}$, 
then, as a consequence of \eqref{intro:gamma-b}, the group $B$ 
is also a residually nilpotent.  Applying formula \eqref{intro:gamma-q} 
we show that an analogous result holds for a split extension of residually 
torsion-free nilpotent groups, provided $C$ acts trivially on 
$A_{\abf}$. Finally, as noted by Bellingeri and Gervais in \cite{BG}, 
formula \eqref{intro:gamma-p} implies that 
split extensions of residually $p$ groups are residually $p$, 
provided $C$ acts trivially on $A_{\ab}\otimes \mathbb{Z}_p$.

We pursue this investigation in \cite{Su-abexact}, where we study the derived series, 
$G \supseteq G'  \supseteq G'' \supseteq \cdots $, and the Alexander invariant, 
$B(G)=G'/G''$, as well as the rational and $p$-versions of these objects, 
together with the cohomology jump loci of a finitely generated group $G$. 
The framework developed here, and the results herein, are used in an 
essential way in \cite{Su-abexact}, as well as in a planned sequel, 
\cite{Su-mfmono}, where we apply these techniques to the study 
of Milnor fibrations of hyperplane arrangements.

\subsection{Organization of the paper} 
\label{intro:organize}
In the first four sections we introduce our main objects of study: 
the $N$-series of a group $G$ and related constructions in 
\S\ref{sect:n-series}; the lower central series $\gamma(G)$ 
and the associated graded Lie algebra $\gr(G)$ in \S\ref{sect:lcs-series}; 
the rational lower central series $\gamma^{\rat}(G)$ 
and its associated graded Lie algebra $\gr^{\rat}(G)$ in \S\ref{sect:lcs-rational};
and finally, the mod-$p$ lower central series $\gamma^p(G)$ and its 
associated graded Lie algebra $\gr^p(G)$ in \S\ref{section:stallings-p}.

The next seven sections constitute the core of the paper. 
We work with split extensions of the form $B=A\rtimes C$; after 
some preparatory material in \S\ref{sect:split}, we 
establish our structural results for $\gamma(B)$ and $\gr(B)$ in 
\S\S\ref{sect:split-lcs}--\ref{sect:ext}; for $\gamma^{\rat}(B)$ 
and $\gr^{\rat}(B)$ in \S\S\ref{sect:rat-split}--\ref{sect:rat-fr}; 
and for $\gamma^p(G)$ and $\gr^p(G)$ in 
\S\S\ref{sect:p-split}--\ref{sect:modp-fr}.

\section{$N$-series}
\label{sect:n-series}   

\subsection{Commutators}
\label{subsec:comms}
Let $G$ be a group.  Given two elements $x,y\in G$, we will write 
${}^x y= xyx^{-1}$, $y^{x}= x^{-1}yx$, and  
$[x,y]=xyx^{-1}y^{-1}$.  The following ``Hall--Witt" 
identities then hold:
\begin{align}
\label{eq:bilg}
& [x , yz]= [x,y] \cdot {}^y [x,z] = [x,y][x,z] [[z,x],y],\\
\label{eq:jacobi}
&[[x,y],{}^y z] [[y,z],{}^z x] [[z,x],{}^x y ]= 1.
\end{align}

Given subgroups $H_1$ and $H_2$ of $G$, define their commutator, 
$[H_1,H_2]$,  to be the subgroup of $G$ generated by all elements 
of the form $[x_1,x_2]$ with $x_i\in H_i$. In particular, 
$G'=[G,G]$ is the derived (or, commutator) subgroup of $G$. 

It is readily seen that $[H_1,H_2]=[H_2,H_1]$, and that $[H_1,H_2]$ 
is contained in $\langle H_1, H_2\rangle$, the subgroup generated by 
$H_1$ and $H_2$. In general, $[H_1,H_2]$ is not contained 
in $H_1$; nevertheless, if  $H_1$ is a normal subgroup, then 
$[H_1,H_2]\subseteq H_1$. Furthermore, $[H_1,H_2]$ need 
not be a normal subgroup, but it is normalized by both $H_1$ 
and $H_2$; if both $H_1$ and $H_2$ are normal subgroups, 
then $[H_1,H_2]$  is also a normal subgroup. Note also that 
$\langle H_1, H_2\rangle=H_1H_2$ precisely when $H_1$ and 
$H_2$ are permuting subgroups, i.e., $H_1H_2=H_2H_1$. 

Very useful in this context is the following ``Three Subgroup Lemma" 
of P.~Hall \cite{Hall}.

\begin{lemma}
\label{lem:3subgroups}
Let $H_1,H_2,H_3$ be three subgroups of $G$, and let $N\triangleleft  G$ 
be a normal subgroup. 
\begin{enumerate}
\item \label{3s1}
If $[[H_1,H_2],H_3]\subseteq N$ and $[[H_2,H_3],H_1]\subseteq N$, then 
$[[H_3,H_1],H_2]\subseteq N$.
\item \label{3s2}
If $H_i\triangleleft  G$, then 
$[[H_1,H_2],H_3]\subseteq [[H_2,H_3],H_1]\cdot [[H_3,H_1],H_2]$.
\end{enumerate}
\end{lemma}

\begin{proof}
Both assertions follow from \eqref{eq:jacobi}; see \cite[Lemma 11.1.6]{Passman} 
and \cite[Theorem 5.2]{MKS}, respectively. 
\end{proof}

The following lemma will also be of much use later on. 
\begin{lemma}
\label{lem:conj-mod}
Let $H_1, H_2$ be subgroups of a group $G$.  Then
$[x,y^k] \equiv [x,y]^k$ modulo $[[H_1,H_2],H_2]$, 
for all $x\in H_1$, $y\in H_2$, and $k\ge 1$.
\end{lemma}

\begin{proof}
The proof is by induction on $k$, with the base case $k=1$ 
tautologically true. By \eqref{eq:bilg}, we have that 
 $[x , yz] = [x,y][x,z] [[z,x],y]$ 
 for every $z\in H_2$. Therefore, 
$[x,yz] \equiv [x,y][x,z] \bmod [[H_1,H_2],H_2]$. 
Taking $z=y^{k-1}$ and using the induction hypothesis gives  
$[x,y^k] \equiv [x,y][x,y^{k-1}]\equiv [x,y]^k \bmod [[H_1,H_2],H_2]$, 
and we are done.
\end{proof}

\subsection{$N$-series}
\label{subsec:n-series}
The following notion is due to Lazard \cite{Lazard}. An {\em $N$-series}\/ 
(or, strongly central series) for a group $G$ is a descending filtration 
\begin{equation}
\label{eq:N-series}
G=K_1 \ge K_2   \ge  \cdots \ge K_n \ge\cdots 
\end{equation}
by subgroups of $G$ satisfying 
\begin{equation}
\label{eq:N-comm}
[K_m, K_n] \subseteq K_{m+n} \: \text{ for 
all $m,n\ge 1$}. 
\end{equation}

In particular, $K=\{K_n\}_{n\ge 1}$ is a {\em central series}, i.e., 
$[G,K_n]\subseteq K_{n+1}$ for all $n\ge 1$, and thus, 
a {\em normal series}, that is, $K_n\triangleleft G$ 
for all $n\ge 1$. Consequently, each quotient $K_n/K_{n+1}$ lies in the center 
of $G/K_{n+1}$, and thus is an abelian group.  If, moreover, all the quotients 
$K_n/K_{n+1}$ (or, equivalently, all the quotients $G/K_{n+1}$) 
are torsion-free, $K$ is called an {\em $N_0$-series}.
 
\subsection{Isolators}
\label{subsec:isolator}
For a subset $S\subseteq G$, we let 
\begin{equation}
\label{eq:isolator}
\sqrt{S}\coloneq \!\sqrt[G]{S}=\{g\in G\mid 
\text{$g^m \in S$ for some $m\in \mathbb{N}$} \}
\end{equation}
be the {\em isolator}\/ (or, {\em root set}) of $S$ in $G$. Clearly, 
$S\subseteq \!\sqrt{S}$ and $\!\sqrt{\!\smash[b]{\!\sqrt{S}}}= \!\sqrt{S}$. 
Moreover, if $\varphi\colon G\to H$ is a homomorphism, and 
$\varphi(S)\subseteq T$, then $\varphi(\!\sqrt[G]{S})\subseteq \!\sqrt[H]{T}$.  

The isolator of a subgroup of $G$ need not be a subgroup; for instance, 
$\!\sqrt[G]{\{1\}}$ is equal to $\Tors(G)$, the set of torsion elements 
in $G$, which is not a subgroup in general. Nevertheless, if $G$ is nilpotent, 
then the isolator of any subgroup of $G$ is again a subgroup. A subgroup 
$H\le G$ is said to be {\em isolated}\/ if $\!\sqrt{H}=H$; clearly, 
$\!\sqrt{H}$ is the intersection of all isolated subgroups of $G$ containg $H$.

Now suppose $N\triangleleft G$ is a normal subgroup. Then 
\begin{equation}
\label{eq:sqrt-normal}
\sqrt[G]{N}=\pi^{-1}(\Tors(G/N))\, ,
\end{equation}
where $\pi\colon G\to G/N$ is the canonical projection, and so 
$\!\sqrt[G]{N}/N\cong \Tors(G/N)$. In particular, 
$N$ is isolated if and only if $G/N$ is torsion-free. 
The following result of Massuyeau (\cite[Lemma 4.4]{Mass}), 
builds on this observation. 

\begin{proposition}[\cite{Mass}]  
\label{prop:sqrt-N}
Suppose $K=\{K_n\}_{n\ge 1}$ is an $N$-series for $G$. Then 
$\!\sqrt{K} \coloneqq \{\!\sqrt{K_n}\}_{n\ge 1}$ is an $N_0$-series 
for $G$.
\end{proposition}

In fact, $\!\sqrt{K}$ is the fastest descending $N_0$-series containing $K$: 
if $L$ is any other $N_0$-series, and $K_n\le L_n$ for all $n\ge 1$, 
then $\!\sqrt{K_n}\le L_n$ for all $n\ge 1$.

\subsection{Dimension series}
\label{subsec:dimension}
For a group $G$ and a commutative ring $R$ with unit $1$, 
we let $R[G]$ be the group $R$-algebra of $G$, and we let 
$I_{R}(G)=\ker (\varepsilon)$ be its augmentation ideal, where 
$\varepsilon  \colon R[G] \to R$ is the augmentation map, 
given by $\varepsilon  \big(\sum r_g g \big)=\sum r_g$. 
A {\em multiplicative filtration}\/ on $I_{R}(G)$ is a descending 
filtration by $R$-submodules, $\Delta=\{\Delta_n\}_{n\ge 1}$, such that 
$\Delta_1=I_{R}(G)$ and $\Delta_m\cdot \Delta_n \subseteq \Delta_{m+n}$ 
for all $m,n\ge 1$.  It follows from the definition that each term 
of the filtration is a two-sided ideal of $R[G]$.  Setting 
\begin{equation}
\label{eq:d-delta}
D^{\Delta}_n=\{ g\in G\mid g-1 \in \Delta_n \}
\end{equation}
defines an $N$-series for $G$, which is an $N_0$-series if $R=\mathbb{Q}$; 
see \cite{Passi79, Passman} and \cite[Lemma 2.4]{Mass}. 

Conversely, any $N$-series $K=\{K_n\}_{n\ge 1}$ 
for $G$ defines a multiplicative filtration, $\{\Delta^K_n\}_{n\ge 1}$, 
on $I(G)$, by setting 
\begin{equation}
\label{eq:delta-k}
\Delta^K_n = \text{$R$-$\spn$} \, \{ (g_1-1)\cdots (g_k-1) \mid  
g_i\in K_{n_i}, \: n_{i_1}+\cdots + n_{i_k}\ge n \} \, ,
\end{equation}
see \cite{Mass}. In turn, such a filtration defines a sequence of 
{\em dimension subgroups},  $D_n(K,R) \le G$;  by the above, 
these subgroups form an $N$-series for $G$. Clearly, $K_n\le D_n(K,R)$ 
for all $n\ge 1$, but the inclusion is strict, in general.

Nevertheless, a classical  result of Malcev, Jennings, and P.~Hall 
insures that $K_n=D_n(K,\mathbb{Q})$ for all $n\ge 1$, provided $K$ is an 
$N_0$-series.  More generally, work of Massuyeau 
(\cite[Theorem 4.1 and Lemma 4.5]{Mass}) shows that 
\begin{equation}
\label{eq:dim-series}
\sqrt{K_n} = D_n(K,\mathbb{Q})= D_n(\!\sqrt{K},\mathbb{Q})\, 
\end{equation}
for all $N$-series $K$ and all $n\ge 1$.

\subsection{The Lie algebra associated to an $N$-series}
\label{subsec:assoc-gr-N}
Let $K=\{K_n\}_{n\ge 1}$ be an $N$-series for a group $G$. 
Since this is a central series, the successive quotients, 
$\gr_{n}^{K} (G)\coloneqq K_n/K_{n+1}$, are abelian groups.  
Following Lazard \cite{Lazard}, set 
\begin{equation}
\label{eq:grPhi}
\gr^{K} (G)  = \bigoplus_{n\ge 1} \gr_{n}^{K} (G). 
\end{equation} 

Using the Witt--Hall identities \eqref{eq:bilg}--\eqref{eq:jacobi} 
and the sassumption that $[K_m,K_n]\subseteq K_{m+n}$ for all 
$m,n\ge 1$, it is readily verified that $\gr^{K} (G)$ has the structure 
of a graded Lie algebra over $\mathbb{Z}$.  The addition is induced from the 
group multiplication, while the Lie bracket, which is induced 
by the group commutator, restricts to bilinear maps 
$[\:,\:]\colon \gr_m^{K}(G)\times  \gr_n^{K}(G)\to 
\gr_{m+n}^{K}(G)$. These operations then satisfy 
the identities 
$[x,x]=0$, $[x,y]+[y,x]=0$, and 
$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$ 
for all homogeneous elements in $\gr^{K}(G)$. 

Now let $G$ and $H$ be two groups endowed with $N$-series 
$K=\{K_n\}_{n\ge 1}$ and $L=\{L_n\}_{n\ge 1}$, respectively, 
and let $\varphi\colon G\to H$ be a group homomorphism 
which is compatible with these filtrations, i.e., 
$\varphi(K_n)\subseteq L_n$ for all $n\ge 1$. 
Then $\varphi$ induces a map, $\gr^{K,L}(\varphi)\colon \gr^K(G)\to \gr^L(H)$.  
It is readily seen that this map preserves degrees and Lie brackets and 
respects compositions;  that is, $\gr^{K,L}(\varphi)$ is a morphism in 
the category of graded Lie algebras.  

\subsection{Dimension series and graded algebras}
\label{subsec:N-alg}

Let $K=\{K_n\}_{n\ge 1}$ be an $N$-series for a group $G$, and  
let $\{\Delta^K_n\}_{n\ge 0}$ be the corresponding filtration on the 
group algebra $R[G]$, where $R$ is a coefficient ring and 
$\Delta^K_0=R[G]$. The associated graded object,
\begin{equation}
\label{eq:gr-kg}
\gr^K(R[G])\coloneqq \bigoplus_{n\ge 0} \Delta^K_n/\Delta^K_{n+1}\, ,
\end{equation}
is an associative algebra, with multiplication induced from the group algebra. 
Upon giving $\gr^K(R[G])$ the Lie bracket $[a,b]=ab-ba$, there is a natural 
map of graded Lie algebras, 
\begin{equation}
\label{eq:gr-map}
\begin{tikzcd}[column sep=18pt]
\gr^K(G)\otimes R \ar[r]& \gr^K(R[G])\, ,
\end{tikzcd}
\end{equation}
defined by $[g]\mapsto [g-1]$ for $g\in K_n$, which extends to a morphism 
from the universal enveloping algebra $U(\gr^K(G,R))$ to the associative 
algebra $\gr^K(R[G])$. When $R=\mathbb{Q}$, Massuyeau \cite{Mass} 
(generalizing a well-known result of Quillen \cite{Quillen68}), 
showed that this morphism is an isomorphism.
 
\section{Lower central series}
\label{sect:lcs-series}

\subsection{The lower central series}
\label{subsec:lcs}

The quintessential example of an $N$-series is the 
{\em lower central series}. For a group $G$, this is the series  
$\gamma(G)=\{\gamma_n(G)\}_{n\ge 1}$, defined inductively by 
$\gamma_1 (G)=G$ and 
\begin{equation}
\label{eq:lcs-filt}
\gamma_{n+1}(G) =[G,\gamma_n (G)].
\end{equation} 
The fact that $\gamma(G)$ is an $N$-series was first established 
by P.~Hall \cite{Hall}, using induction and Lemma~\ref{lem:3subgroups}; 
see also \cite{MKS, Passman, Serre}.    
As the next, well-known lemma shows, $\gamma(G)$ is the 
fastest descending central series of $G$, and thus, the fastest 
descending $N$-series of $G$. 

\begin{lemma}
\label{lem:fastest-n}
If $K$ is a descending central series for $G$, then $\gamma_n(G)\le K_n$ 
for all $n\ge 1$. 
\end{lemma}

\begin{proof}
Induction on $n$, with the base case $n=1$ being obvious: assuming 
$\gamma_{n}(G)\subseteq K_{n}$, we have that 
$\gamma_{n+1}(G)=[G,\gamma_{n}(G)] \subseteq [G,K_{n}]
\subseteq K_{n+1}$, and we are done. 
\end{proof}

The successive quotient of the series, $\gamma_n(G)/\gamma_{n+1}(G)$, 
are abelian groups. The first such quotient, $G/\gamma_{2}(G)$, coincides  
with the abelianization $G_{\ab}=H_1(G,\mathbb{Z})$.  If $\varphi\colon G\to H$ 
is a group homomorphism, an easy induction  shows that 
$\varphi(\gamma_n(G))= \gamma_n(\varphi(G))$, for all $n\ge 1$. 
Thus, $\varphi(\gamma_n(G))\subseteq \gamma_n(H)$, with equality 
if $\varphi$ is surjective.

By definition, $G$ is {\em nilpotent}\/ if $\gamma(G)$ terminates in 
finitely many steps.  For each $n\ge 1$, the quotient group 
$\Gamma_n= G/\gamma_n(G)$ is a nilpotent group, to wit, 
the maximal $(n-1)$-step nilpotent quotient of $G$. 
Since this group is nilpotent, its torsion elements, 
$\Tors(\Gamma_n)$, form a (characteristic) subgroup; the quotient
group, $\Gamma_n/\Tors(\Gamma_n)$, is the  maximal 
$(n-1)$-step torsion-free nilpotent quotient of $G$. 

\subsection{Further properties}
\label{subsec:further-lcs}
The central result relating group homology to the lower central series  
is the following theorem of Stallings \cite[Theorem 3.4]{St}; 
for significant refinements of this theorem, we 
refer to Dwyer \cite{Dwyer} and Cochran--Harvey \cite{CH-gt08}.

\begin{theorem}[\cite{St}]
\label{thm:stallings}
Let $\varphi\colon A\to B$ be a homomorphism that induces an isomorphism 
$H_1(A,\mathbb{Z})\isom H_1(B,\mathbb{Z})$ and an epimorphism $H_2(A,\mathbb{Z})\twoheadrightarrow H_2(B,\mathbb{Z})$. 
Then $\varphi$ induces isomorphisms $A/\gamma_n(A) \isom B/\gamma_n(B)$ 
for all $n\ge 1$.
\end{theorem}

The lower central series and other related series control 
certain residual properties of groups. A group $G$ is said to 
be \emph{residually $\mathcal{P}$},  where $\mathcal{P}$ is a class of groups, 
if for any $g\in G$, $g\ne 1$, there exists a group $Q\in \mathcal{P}$ 
and an epimorphism $\psi\colon G\to Q$ such that $\psi(g)\ne 1$. 
The following lemma is well-known, and easy to prove.

\begin{lemma}
\label{lem:res-nilp}
A group $G$ is residually nilpotent if and only if the intersection 
of its lower central series, 
$\gamma_{\omega}(G)\coloneqq \bigcap_{n\ge 1} \gamma_{n}(G)$, 
is the trivial subgroup.
\end{lemma}

\subsection{The associated graded Lie algebra of a group}
\label{subsec:assoc-gr}
Returning to the setup from \S\ref{subsec:assoc-gr-N}, 
consider the case when the $N$-series $K$ is $\gamma(G)$,  
the lower central series of $G$. The resulting Lie algebra,  $\gr^K(G)$, 
denoted simply by $\gr(G)$, is called the {\em associated 
graded Lie algebra}\/ of $G$ (over the ring $\mathbb{Z}$).

Since $\gamma_{n+1}(G)=[G,\gamma_{n}(G)]$, the Lie bracket map 
$[\:,\:]\colon \gr_1(G) \otimes \gr_{n}(G)\to \gr_{n+1}(G)$ is surjective. 
It follows by induction on $n$ that $\gr(G)$ is generated as a Lie algebra 
by its degree $1$ piece, $\gr_1(G)=G_{\ab}$.  An inductive argument 
now shows the following: if $G_{\ab}$ is finitely generated, then the 
groups $\gr_n(G)$ are also finitely generated, for all $n\ge 1$.

\begin{example}
\label{ex:free}
Let $F_X$ be the free group on a set $X$, and let $\Lie(X)$ be the 
free Lie algebra on this set. Work of P.~Hall \cite{Hall}, 
W.~Magnus  \cite{Magnus35, Magnus40}, 
 and E.~Witt from the 1930s (see \cite{MKS,Serre}) shows 
that the canonical map $X\to \gr_1(F_X)$ induces an isomorphism 
of graded Lie algebras, $\Lie(X)\isom \gr(F_X)$. Consequently, the 
groups $\gr_n(F_X)$ are torsion-free; moreover, in the case when 
$X$ is finite, those groups have rank equal to 
$\tfrac{1}{n}\sum_{d\mid n} \mu(d) \abs{X}^{n/d}$, 
where $\mu\colon \mathbb{N}\to \{0,\pm 1\}$ is the M\"{o}bius function.
\end{example}

Now let $\varphi\colon G\to H$ be a group homomorphism. 
As noted in \S\ref{subsec:lcs}, we have 
$\varphi(\gamma_n(G))= \gamma_n(\varphi(G))$ for all $n\ge 1$;   
it follows that $\varphi$ induces a morphism of graded Lie algebras, 
$\gr(\varphi)\colon \gr(G)\to \gr(H)$.  It is readily seen that 
the assignment $\varphi \leadsto \gr(\varphi)$ is functorial;  
moreover, if $\varphi$ is surjective, then $\gr(\varphi)$ 
is also surjective.

Given any $N$-series $K=\{K_n\}_{n\ge 1}$, recall that we have 
inclusions $\gamma_n (G)\subseteq K_n$ for all $n\ge 1$.  
Consequently, we have an induced morphism of graded 
Lie algebras, $\gr(G) \to \gr^{K} (G)$. 
In degree $1$, this map  is surjective, but in higher 
degrees it may fail to be either injective or surjective.

\section{Rational lower central series}
\label{sect:lcs-rational}

\subsection{The rational lower central series}
\label{subsec:lcs-rat}

The rational version of the lower central series was introduced 
by Stallings in \cite{St} (see also \cite{CH-gt08, MP}). For a group $G$, 
this is the series $\gamma^{\rat}(G)=\{\gamma^{\rat}_{n}(G)\}_{n\ge 1}$, 
defined inductively by $\gamma^{\rat}_{1} (G)=G$ and 
\begin{equation}
\label{eq:gamma-q-filtration}
\gamma^{\rat}_{n+1}(G)=\!\sqrt{ [G,\gamma^{\rat}_{n}(G)]}\, .
\end{equation}  

As observed in \cite{St}, $\gamma^{\rat}(G)$ is the most rapidly 
descending central series whose successive quotients are torsion-free 
abelian groups; in particular, it is the fastest descending $N_0$-series 
for $G$.  Clearly, the terms of this series are fully 
invariant subgroups of $G$, i.e., $\varphi(\gamma^{\rat}_{n}(G))\subseteq 
\gamma^{\rat}_{n}(H)$ for every homomorphism $\varphi\colon G\to H$. 
The next proposition gives another, very useful description of this series.

\begin{proposition}
\label{prop:sq-bar}
$\gamma^{\rat}_{n}(G)= \!\sqrt{\gamma_n(G)}$, for all $n\ge 1$.
\end{proposition}

\begin{proof}
First we show that $\!\sqrt{\gamma_n(G)} \subseteq \gamma^{\rat}_{n}(G)$ for all $n$.  
We use induction on $n$, the base case $n=1$ being clear. For the induction step, 
we have
\[
\!\sqrt{\gamma_{n+1}(G)}=\!\sqrt{[G,\gamma_{n}(G)]} 
\subseteq \!\sqrt{\Big[G,\!\sqrt{\gamma_{n}(G)}\Big]} 
\subseteq \!\sqrt{\big[G,\gamma^{\rat}_{n}(G)\big] } 
=\gamma^{\rat}_{n+1}(G) \, .
\]

For the reverse inclusion, $\gamma^{\rat}_{n}(G) \subseteq \!\sqrt{\gamma_{n}(G)}$, 
we use again induction on $n$.  For the induction step, we have
\[
\gamma^{\rat}_{n+1}(G) =\!\sqrt{\big[G,\gamma^{\rat}_{n}(G)\big]} 
\subseteq \!\sqrt{\Big[G,\!\sqrt{\gamma_{n}(G)}\Big]} 
\subseteq \!\sqrt{\!\!\sqrt{\gamma_{n+1}(G)}} 
=\!\sqrt{\gamma_{n+1}(G)}\, ,
\]
where at the third step we used that, as a consequence of 
Proposition \ref{prop:sqrt-N}, $\!\sqrt{\gamma(G)}$ is a 
normal series. This completes the proof.
\end{proof}

It follows from \eqref{eq:gamma-q-filtration} and the 
above proposition that $\!\sqrt{\gamma_{n+1}(G)}=
\!\sqrt{ \big[G,\!\sqrt{\gamma_{n}(G)}\, \big]}$\, for all $n\ge 1$.

\begin{corollary}
\label{cor:stallings-q}
The series $\gamma^{\rat}(G)$ is an $N_0$-series. 
\end{corollary}

\begin{proof}
By \cite[Lemma 1.3]{Passi79} and \cite[Lemma 11.1.8]{Passman}, 
or as a corollary to Proposition \ref{prop:sqrt-N}, 
we have that $\!\sqrt{\gamma(G)}$ is an $N_0$-series. 
The claim now follows from Proposition  \ref{prop:sq-bar}.
\end{proof}

Note that $G/\gamma^{\rat}_{2}(G)=G_{\abf}$, where 
$G_{\abf}=G_{\ab}/\Tors(G_{\ab})$ is the maximal 
torsion-free abelian quotient of $G$. More generally, 
as pointed out in  \cite[Appendix A]{FHT}, 
\begin{equation}
\label{eq:g-mod-tau}
G/\gamma^{\rat}_{n}(G)= \Gamma_n /\Tors(\Gamma_n),
\end{equation}
where $\Gamma_n=G/\gamma_n(G)$, and this property defines the 
series $\gamma^{\rat}(G)$.  Consequently, 
the quotients $G/\gamma^{\rat}_{n}(G)$ are torsion-free nilpotent 
groups, and the series $\gamma^{\rat}(G)$ is the fastest descending 
normal series with this property. Furthermore, it follows from 
\eqref{eq:sqrt-normal} that 
$\gamma^{\rat}_{n}(G)/\gamma_{n}(G)\cong \Tors(\Gamma_n)$.

\subsection{Alternate definitions}
\label{subsec:Bass-Lubotzky}
We now describe several other ways in which one can define series 
which turn out to coincide the rational lower central series. The first 
one is the series $\gamma^{\zz} (G)=\{\gamma^{\zz}_n (G)\}_{n\ge 1}$, 
defined inductively by Bass and Lubotzky in \cite[Definition 7.1]{BL} 
by means of the following three conditions:
\begin{enumerate}
\item $\gamma_1^{\zz}(G)=G$;
\item $[G,\gamma_n^{\zz}(G)] \le \gamma_{n+1}^{\zz}(G)$;
\item $\gamma_{n+1}^{\zz}(G)/[G,\gamma_n^{\zz}(G)] =
\Tors\!\big(\gamma_{n}^{\zz}(G)/[G,\gamma_n^{\zz}(G)] \big)$.
\end{enumerate}

The second condition implies that $\gamma^{\zz}_n(G)\triangleleft G$, 
while the third one is equivalent to 
$\gamma_n^{\zz}(G)/\gamma_{n+1}^{\zz}(G)=(\Gamma_n^{\zz})_{\abf}$, 
where $\Gamma_n^{\zz}=\gamma_{n}^{\zz}(G)/[G,\gamma_n^{\zz}(G)] $. 
As shown in \cite[Proposition 7.2(b)]{BL}, for each $n\ge 1$ there is 
an exact sequence 
\begin{equation}
\label{eq:bass-lubotzky-seq}
\begin{tikzcd}[column sep=20pt]
1\ar[r]& \gamma^{\zz}_n(G)/\gamma_n(G) \ar[r]& \Gamma_n  
\ar[r]& G/\gamma^{\zz}_n (G) \ar[r]& 1\, ,
\end{tikzcd}
\end{equation} 
where $\gamma^{\zz}_n (G)/\gamma_n (G)= \Tors(\Gamma_n)$. 
It follows that the successive quotients of the series $\gamma^{\zz}(G)$
are torsion-free, and the natural maps 
$\gamma_n(G)/\gamma_{n+1}(G)\to \gamma^{\zz}_n(G)/\gamma^{\zz}_{n+1}(G)$ 
have finite kernel and cokernel. 
For $n=1$, we have that  $\gamma^{\zz}_1 (G)/\gamma^{\zz}_2 (G)=G_{\abf}$, and the 
previous map is simply the projection $G_{\ab}\twoheadrightarrow G_{\abf}$. 
Finally, as noted in \cite{BL}, $\gamma^{\zz} (G)$ 
is the fastest descending central series whose successive quotients are 
torsion-free.  Therefore, $\gamma^{\zz}_{n}(G)= \gamma^{\rat}_n(G)$, 
for all $n\ge 1$.

Closely related is the descending series $\gamma^T(G)=\{\gamma^{T}_n (G)\}_{n\ge 1}$,  
defined by Koberda \cite{Ko} by setting  $\gamma^{T}_1 (G) =G$ and 
$\gamma^T_{n} (G)=\ker \big(G\twoheadrightarrow \Gamma_n  
\twoheadrightarrow \Gamma_n/\Tors(\Gamma_n)\big)$.
It follows from either \eqref{eq:g-mod-tau} or \eqref{eq:bass-lubotzky-seq} that 
$\gamma^{T}_n(G)= \gamma^{\rat}_{n} (G)= \gamma^{\zz}_{n} (G)$ 
for all $n$.

Finally, another description is given in \cite[\S 10.4]{Hillman}, where 
Hillman defines inductively a sequence of subgroups, $\{G_{n}^{\rat}\}_{n\ge 1}$, 
as follows: $G_1^{\rat}=G$ and $G_{n+1}^{\rat}$ is the preimage 
in $G$ of $\Tors(G/[G,G_{n}^{\rat}])$. Then all the quotients $G/G_n^{\rat}$ 
are torsion-free nilpotent groups, and the series $\{G_{n}^{\rat}\}_{n\ge 1}$ 
is the fastest descending normal series with this property. Consequently, 
$G_{n}^{\rat}=\gamma_{n}^{\rat}(G)$ for all $n$.

\subsection{Further properties}
\label{subsec:further-q}

In \cite[Theorem7.3]{St}, Stallings gave a rational analogue of 
his celebrated result relating homology to nilpotent quotients. 
We record this theorem next; 
for more elaborate versions, we refer 
to \cite{Dwyer, CH-gt08}.

\begin{theorem}[\cite{St}]
\label{thm:stallings-rat}
Let $\varphi\colon A\to B$ be a homomorphism that induces an isomorphism 
$H_1(A,\mathbb{Q})\isom H_1(B,\mathbb{Q})$ and an epimorphism $H_2(A,\mathbb{Q})\twoheadrightarrow H_2(B,\mathbb{Q})$. 
Then $\varphi$ induces monomorphisms 
$A/\gamma^{\rat}_n(A) \hookrightarrow B/\gamma^{\rat}_n(B)$ 
and isomorphisms 
$\big(\gamma^{\rat}_n(A)/\gamma^{\rat}_{n+1}(A) \big)\otimes \mathbb{Q}
\isom \big( \gamma^{\rat}_n(B)/\gamma^{\rat}_{n+1}(B) \big)\otimes \mathbb{Q}$\, 
for all $n\ge 1$.
\end{theorem}

A group $G$ is residually torsion-free nilpotent (RTFN) if every 
non-trivial element can be detected in a torsion-free nilpotent 
quotient.

\begin{lemma}
\label{lem:rtfn}
A group $G$ is RTFN  if and only if the intersection of its rational 
lower central series, 
$\gamma^{\rat}_{\omega}(G)\coloneqq \bigcap_{n\ge 1} \gamma^{\rat}_{n}(G)$, 
is the trivial subgroup.
\end{lemma}

A proof of this well-known result is given in \cite[Proposition 7.2(e)]{BL}. 
By \cite[Ch.~VI, Thm.~2.26]{Passi79}, $G$ is RTFN precisely when the 
group-algebra $\mathbb{Q}[G]$ is residually nilpotent, that is, 
$\bigcap_{n\geq 1}I_{\mathbb{Q}}(G)^{n}=\{0\}$.

Residually nilpotent groups are residually nilpotent (and therefore 
residually finite), residually $p$ for every prime $p$ 
(see \S\ref{subsec:more-p-stallings}), and bi-orderable. 
If $G$ is residually nilpotent and $\gr_{n} (G)$ is torsion-free for 
$n\ge 1$, then $G$ is residually torsion-free nilpotent. 
Examples of RTFN groups include torsion-free nilpotent 
groups, free groups, orientable surface groups, and 
right-angled Artin groups.

\begin{remark}
\label{rem:dim-rat}
Let $R[G]$ be group algebra of a group $G$ with coefficients in a ring $R$, and 
let $I_{R}(G)$ be its augmentation ideal. Clearly, the powers of this ideal, 
$\{I_{R}(G)^n\}_{n\ge 1}$, form a multiplicative filtration of $I(G)$. 
By \eqref{eq:d-delta}, this filtration defines a sequence of dimension 
subgroups, $\{D_n(G,R)\}_{n\ge 1}$, which forms an $N$-series 
for $G$. We then have $\gamma_n(G) \le D_n(G,R)$, 
but the inclusion is proper in general. By \eqref{eq:dim-series}, 
the equality $\!\sqrt{\gamma_n(G)}=D_n(G,\mathbb{Q})$ holds for all $n\ge 1$;  
for other proofs, we refer to \cite[Lemma 1.4]{Passi79} and 
\cite[Theorem 11.1.10]{Passman}.
\end{remark}

\subsection{The rational associated graded Lie algebra}
\label{subsec:lcs-stallings-gr}
Recall that the successive quotients of the rational lower central series, 
$\gr_n^{\rat}(G)=\gamma^{\rat}_n (G)/ \gamma^{\rat}_{n+1} (G)$,  
are torsion-free abelian groups.  The direct sum of these quotients, 
\begin{equation}
\label{eq:grg-q}
\gr^{\rat}(G)\coloneqq \bigoplus\nolimits_{n\ge 1} 
\gamma^{\rat}_{n} (G)/ \gamma^{\rat}_{n+1} (G),
\end{equation}
with Lie bracket induced from the group commutator, 
is the {\em rational associated graded Lie algebra}\/ of $G$. 
Since the terms of $\gamma^{\rat}(G)$ are fully invariant 
subgroups of $G$, this construction is again functorial.

Since $\gamma_n(G)\le \gamma^{\rat}_n(G)$ for all $n$, we 
have an induced map between associated graded Lie 
algebras, $\gr(G)\to \gr^{\rat}(G)$, which is natural with respect 
to group homomorphisms. In \cite[Proposition 7.2]{BL}, Bass and 
Lubotzky show that, for each $n\ge 1$, there is an exact sequence,
\begin{equation}
\label{eq:bass-lubotzky-seq-bis}
\begin{tikzcd}[column sep=17pt]
1\ar[r]& \displaystyle{\frac{\gamma_n(G) \cap\gamma^{\rat}_{n+1}(G)}{\gamma_{n+1}(G)} }
\ar[r] &\gr_n(G)  \ar[r]& \gr^{\rat}_n(G)  \ar[r]
&\displaystyle{\frac{\gamma^{\rat}_n(G)}{\gamma_n(G) \gamma^{\rat}_{n+1}(G) \ar[r]}}& 1\, .
\end{tikzcd}
\end{equation} 
Since the groups on the left and the right are both finite, 
the next result follows.

\begin{proposition}[\cite{BL}]
\label{prop:bl-grlie}
For a group $G$, the following hold.
\begin{enumerate}[itemsep=2pt]
\item \label{bl-gr1}
The map $\gr(G)\to \gr^{\rat}(G)$ has torsion kernel 
and cokernel in each degree.

\item \label{bl-gr2}
The map $\gr(G)\otimes \mathbb{Q}\to \gr^{\rat}(G)\otimes \mathbb{Q}$ 
is an isomorphism.
\end{enumerate}
\end{proposition}

As a consequence, the rational Lie algebra $\gr^{\rat}(G)\otimes \mathbb{Q}$ 
is generated in degree $1$ by the rational vector space 
$G_{\ab}\otimes \mathbb{Q}=H_1(G,\mathbb{Q})$. 

\section{Mod-$p$ lower central series}
\label{section:stallings-p}

\subsection{$p$-Torsion series and $N_p$-series}
\label{subsec:p-torsion-np}
Fix a prime $p$.  For a subset $S$ of a group $G$, we let 
$S^p$ be the subgroup generated by $\{g^p \mid g\in G\}$. 
Clearly,  if $\varphi \colon G\to H$ is a homomorphism, then 
$\varphi(S^p)\subseteq (\varphi(S))^p$.

A sequence $K=\{K_n\}_{n\ge 1}$ of subgroups of $G$ is said to be 
a {\em $p$-torsion series}\/ (or, {\em $p$-central filtration}) if 
$K_n^p\subseteq K_{n+1}$ for all $n\ge 1$, see~\cite{BL,Darne,HM}. 
Now suppose that $K$ is also an $N$-series. 
Then the successive quotients, $\gr_n^{K}(G)=K_n/K_{n+1}$,  
are $p$-torsion abelian groups, that is, elementary abelian $p$-groups. 
Hence, the groups $\gr_n^{K}(G)$ can be viewed as $\mathbb{Z}_p$-vector spaces, 
and the associated graded Lie algebra, $\gr^K(G)=\bigoplus_{n\ge 1} \gr_n^{K}(G)$, 
becomes a Lie algebra over the prime field $\mathbb{Z}_p$. 

A sequence $K=\{K_n\}_{n\ge 1}$ of subgroups of $G$ is said to be an 
{\em $N_p$-series} (or, a $p$-restricted series) if $K$ is an $N$-series and 
$K_n^p\subseteq K_{pn}$ for all $n\ge 1$. An $N_p$-series is 
both an $N$-series and a $p$-torsion series, but the converse does 
not hold.  To every $N$-series $K$ and prime $p$, 
there corresponds a canonical $N_p$-series, $K^{[p]}$; 
its terms are given by
\begin{equation}
\label{eq:np-series}
K^{[p]}_{n}=\prod_{\substack{mp^{j}\ge n\\m\ge 1, j\ge 0}} \big(K_m \big)^{p^j} .
\end{equation}
This is the fastest descending $N_p$-series containing $K$, cf.~\cite{HM}.

As shown by Lazard in \cite[Corollary 6.8]{Lazard}, 
the associated graded Lie algebra of an $N_p$-series $K$ 
is a {\em $p$-restricted Lie algebra}, in the sense of Jacobson \cite{Jac}.
In addition to the usual Lie bracket, such a Lie algebra, $\gr^{K}(G)$, comes 
endowed with a $p$-th power map, $\gr^{K}_n(G)\to \gr^{K}_{n+1}(G)$, 
which is induced from the map $G\to G$, $g\mapsto g^p$; moreover, the two 
operations satisfy the compatibility conditions listed in \cite{Jac}, 
see \cite{Darne, HM}.  

\subsection{The mod-$p$ lower central series}
\label{subsec:lcs-stallings-p}
In \cite{St}, Stallings introduced the {\em mod-$p$ lower central series}, 
a subgroup series which has been much studied since then, 
see for instance \cite{BL,BG,CH-jlms08,Cooper,Darne,HM,Paris}. 
Given a group $G$, this series, denoted 
$\gamma^{p}(G)=\{\gamma^{p}_{n}(G)\}_{n\ge 1}$, 
is defined inductively by setting  $\gamma^{p}_{1} (G)=G$ and 
\begin{equation}
\label{eq:gamma-p-filtration}
\gamma^{p}_{n+1}(G)=\left\langle\big(\gamma^{p}_{n}(G)\big)^p ,
\left[G,\gamma^{p}_{n}(G)\right] \right\rangle .
\end{equation} 

By construction, the terms of this series 
satisfy $\big(\gamma^{p}_{n}(G)\big)^p\subseteq \gamma^{p}_{n+1}(G)$; 
thus, $\gamma^{p}(G)$ is a $p$-torsion series. Clearly, the terms of  
$\gamma^{p}(G)$  are fully invariant subgroups of $G$. Note that 
$\gamma^{p}_{2} (G)=\langle G^p, G'\rangle$, and so 
$G/\gamma^{p}_{2} (G)=G_{\ab}\otimes \mathbb{Z}_p = H_1(G,\mathbb{Z}_p)$. 

\begin{lemma}[\cite{Paris}]
\label{lem:paris}
$\gamma^{p}(G)$ is an $N$-series for $G$.
\end{lemma}

Consequently, $\gamma^{p}(G)$ is a descending central series.  In fact, 
this is the fastest descending such series among all $p$-torsion 
series for $G$; for completeness, we give a quick proof of this known fact. 

\begin{lemma}
\label{lem:fastest-p}
If $K$ is a descending, central, $p$-torsion series for $G$, then 
$\gamma^{p}_n(G)\le K_n$ for all $n\ge 1$. 
\end{lemma}

\begin{proof}
The proof is by induction on $n$, with the base case 
$n=1$ being obvious.  Assuming $\gamma^{p}_n(G)\subseteq K_{n}$, 
we have 
$\gamma^{p}_{n+1}(G)=\left\langle\big(\gamma^{p}_n(G)\big)^p, 
[G,\gamma^{p}_n(G)] \right\rangle
\subseteq  \left\langle(K_n)^p, [G,K_n] \right\rangle
\subseteq K_{n+1}$, 
and we are done. 
\end{proof}

The Stallings series $\gamma^{p}(G)$ is not in general an 
$N_p$-series. Rather, the canonical $N_p$-series associated to the 
lower central series, $\gamma^{[p]}(G)$, coincides with the series 
introduced by H.~Zassenhaus in 1939, see \cite{Cooper, Darne, HM}.  
The two filtrations are cofinal (and thus define the same topology 
on $G$), but they may differ quite a lot at a granular level. 
For instance, if $G$ is abelian, then $\gamma^{p}_n(G)=G^{p^{n-1}}$, 
whereas $\gamma^{[p]}_n(G)=G^{p^{j}}$, where $j=\lceil \log_p(n) \rceil$. 

\subsection{Associated graded Lie algebras}
\label{subsec:gr-stallings-p}
Since both $\gamma^{p}(G)$ and $\gamma^{[p]}(G)$ are 
$p$-torsion $N$-series, the corresponding associated graded 
Lie algebras, $\gr^{p}(G)$ and $\gr^{[p]}(G)$, are Lie algebras 
over $\mathbb{Z}_p$.  
Clearly, the terms of both series are fully invariant 
subgroups of $G$; thus the constructions $G\leadsto \gr^p(G)$ 
and $G\leadsto \gr^{[p]}(G)$ are both functorial. 

\begin{example}
\label{ex:z-p}
For the cyclic group $\mathbb{Z}$, we have 
$\gr^{p}_n(\mathbb{Z})=\mathbb{Z}_p$, for all $n\ge 1$. On the other hand, 
$\gr^{[p]}_n(\mathbb{Z})$ is equal to $\mathbb{Z}_p$ if $n$ is a non-negative 
power of $p$, and equal to $0$, otherwise.
\end{example}

We will focus in \S\ref{sect:p-split} on $\gr^p(G)$, 
the {\em mod-$p$ associated graded Lie algebra}\/ of $G$. 
For now, let us make an observation. 
The power-$p$ map $G\to G$, $x\mapsto x^p$ sends 
each subgroup $\gamma^p_n(G)$ to $\gamma^p_{n+1}(G)$, 
and thus defines maps $\gr^p_n(G) \to \gr^p_{n+1}(G)$ 
by passing to quotients. As noted in \cite[\S{12}]{BL}, 
the $\mathbb{Z}_p$-Lie algebra $\gr^p(G)$ is generated---through   
Lie brackets and these power operations---by its degree 
$1$ piece, $\gr^p_1(G)=H_1(G,\mathbb{Z}_p)$.

\subsection{Further properties}
\label{subsec:more-p-stallings}
In \cite[Theorem~3.4]{St}, Stallings proved a mod-$p$ analogue of his 
celebrated result relating homology and lower central series; for a more refined 
version of this theorem, we refer to \cite{CH-jlms08}.

\begin{theorem}[\cite{St}]
\label{thm:stallings-p}
Let $\varphi\colon A\to B$ be a homomorphism that induces an isomorphism 
$H_1(A,\mathbb{Z}_p)\isom H_1(B,\mathbb{Z})$ and an epimorphism $H_2(A,\mathbb{Z}_p)\twoheadrightarrow H_2(B,\mathbb{Z}_p)$. 
Then $\varphi$ induces isomorphisms $A/\gamma^{p}_n(A) \isom B/\gamma^{p}_n(B)$ 
for all $n\ge 1$.
\end{theorem}

A group $G$ is said to be {\em residually $p$}\/ if every nontrivial element 
of $G$ can be detected in a finite $p$-group quotient. Residually $p$ groups 
are residually nilpotent and residually finite. As shown in 
\cite[Proposition 2.3]{Paris}, this property is detected by the 
intersection of the mod-$p$ lower central series, 
$\gamma^p_{\omega}(G)\coloneqq \bigcap_{n\ge 1} \gamma^{p}_{n} (G)$.

\begin{lemma}[\cite{Paris}] 
\label{lem:paris-bis}
Let $G$ be a finitely generated group. Then $G$ is residually $p$ if and 
only if\/ $\gamma^p_{\omega}(G)=\{1\}$.
\end{lemma}

\section{Split exact sequences and commutators}
\label{sect:split}

\subsection{Semidirect products}
\label{subsec:semi}
We now switch our focus, from arbitrary groups  to 
those that arise as a semi-direct product of two other groups. 
To get started, consider a split exact sequence of groups, 
\begin{equation}
\label{eq:abc-exact}
\begin{tikzcd}[column sep=24pt]
1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] \arrow[bend right=45,"\sigma",swap]{l} & 1\, ,
\end{tikzcd}
\end{equation}
where $\beta\circ\sigma=\id_C$. Let $\Aut(A)$ be the group of 
automorphisms of $A$, with group operation 
$\psi_1\cdot \psi_2 = \psi_1\circ \psi_2$. The splitting homomorphism 
$\sigma$ defines an action of $C$ on $A$ via the homomorphism 
$\varphi \colon C\to \Aut(A)$ given by 
\begin{equation}
\label{eq:action}
\alpha(\varphi(c)(a))=\sigma(c) \alpha(a) \sigma(c)^{-1}.
\end{equation}

This procedure realizes the group $B$ as a split extension, $B=A\rtimes_{\varphi} C$; 
that is, the set $A\times C$, endowed with the group operation given by
\begin{equation}
\label{eq:gp-op}
(a_1,c_1)\cdot (a_2,c_2)=(a_1\varphi(c_1)(a_2) , c_1c_2)\, .
\end{equation}
Conversely, every split extension $B=A\rtimes_{\varphi} C$ 
gives rise to a split exact sequence of the form \eqref{eq:abc-exact} by 
defining $\alpha(a)=(a,1)$, $\beta(a,c)=c$, and $\sigma(c)=(1,c)$.  

In what follows, we will identify the group $C$ with its image under the 
splitting $\sigma$, and thus view $C$ as a subgroup of 
$B=A\rtimes_{\varphi} C$. Likewise, 
we will identify $A$ with its image under the inclusion $\alpha$ and 
view it as a normal subgroup of $B$.  With these identifications, 
the action of $C$ on $A$ becomes the restriction of the conjugation 
action in $B$, that is, 
\begin{equation}
\label{eq:comm-act}
\varphi(c)(a)=cac^{-1} .
\end{equation}
Furthermore, every element $b\in B$ can be written in a unique 
way as a product, $b=ac$, for some $a\in A$ and $c\in C$.

\subsection{Commutators and powers}
\label{subsec:comm-semi}
Observe that $\varphi(c)(a)\cdot a^{-1}=[c,a]$ and 
$a\cdot \varphi(c)(a^{-1})=[a,c]=[c,a]^{-1}$. 
Let us also note the way $C$ acts on two types of 
commutators: if $g, c\in C$ and $a, a'\in A$, then
\begin{align}
\label{eq:act-1}
\varphi(g)([a,a'])&=[\varphi(g)(a),\varphi(g)(a')] \, ,
\\ 
\label{eq:act-2}
\varphi(g)([c,a])&=
[gc,a]\cdot  {}^a[g,a^{-1}]  \, .
\end{align}

Moreover, conjugation by an element $x\in A$ 
acts on a commutator $[c,a]$ as follows:
\begin{align}
\label{eq:conj-1}
x [c,a] x^{-1}&= [c, x^c a] [c,x^c]^{-1}
\\ 
\label{eq:conj-2}
&=[c, {}^{x^c\!}a] [x^c,a] [a,x]\,.
\end{align}

We conclude this section with two lemmas that will be needed later on. 
In both lemmas, $B=A\rtimes C$ is a semidirect product of groups.

\begin{lemma}
\label{lem:normal}
Let $A_1\le A$ and $C_1\le C$, and assume $[A_1,C_1]\le A_1$. 
Then $[A_1,C_1]$ is a normal subgroup of $A_1$. In particular, 
$[A,C_1]\triangleleft A$ and $[A,C]\triangleleft A$.
\end{lemma}

\begin{proof}
By our assumption, if $c\in C_1$ and $x\in A_1$, then 
$x^c = [c^{-1},x] x\in A_1$. Thus, if $a\in A_1$, then  
$x [c,a] x^{-1}$ belongs to $[A_1,C_1]$, by formula \eqref{eq:conj-1}. 
The claim and its consequences follow at once.
\end{proof}

\begin{lemma}
\label{lem:power}
Let $A_1\le  A$ and $C_1\le C$, and assume $[A_1,C_1]\le A_1$. 
Then $(ac)^k \equiv a^kc^k \bmod [A_1,C_1]$, for all 
$a\in A_1$, $c\in C_1$, and $k\ge 1$.
\end{lemma}

\begin{proof}
We prove the claim by induction on $k$, with the base case $k=1$ 
being obvious. Suppose $(ac)^{k-1} \equiv a^{k-1}c^{k-1}$ 
modulo $[A_1,C_1]$, that is, $(ac)^{k-1} = a^{k-1} a' c^{k-1}$, 
for some $a'\in [A_1,C_1]$. Then 
\[
(ac)^{k} =  a^{k-1} a' c^{k-1} ac= 
a^k \cdot [a^{-1}a',c^{k-1}][c^{k-1},a^{-1}a'a] \cdot a^{-1}a'a \cdot c^k \, .
\]
But $a^{-1}a'a\in [A_1,C_1]$ by Lemma \ref{lem:normal}, and we are done.
\end{proof}

\section{The lower central series of a split extension}
\label{sect:split-lcs}

\subsection{The Guaschi--Pereiro series}
\label{subsec:split-n-series}

Our goal in this section is to analyze the lower central series 
$\gamma(B)=\{\gamma_n(B)\}_{n\ge 1}$ of a split extension of groups, 
$B=A\rtimes_{\varphi} C$, and describe it in terms of the 
corresponding lower central series for the factors, $\gamma(A)$ and 
$\gamma(C)$. In \cite{GP}, Guaschi and Pereiro associate to such 
a split extension a sequence of subgroups of $A$ which plays 
a central role in this analysis. 

This sequence, $L=\{L_n\}_{n\ge 1}$, is defined inductively 
by setting $L_1=A$ and letting $L_{n+1}$ be the subgroup 
of $A$ generated by the commutators $[A,L_{n}]$, 
$[A, \gamma_{n}(C)]$,  and $[L_{n},C]$; that is, 
\begin{equation}
\label{eq:subgroup}
L_{n+1}=\langle [A,L_{n}], [A, \gamma_{n}(C)], [L_{n},C]\rangle \, .
\end{equation}
For instance, 
$L_2=\langle A', [A,C]\rangle$ and 
$L_3=\langle \gamma_3(A), [A,C'],[A',C], [[A,C],A], [[A,C],C]\rangle$.
Let us emphasize that this sequence depends in an essential way on the 
monodromy $\varphi\colon C\to \Aut(A)$ of the split 
extension $B=A\rtimes_{\varphi} C$. 

\begin{remark}
\label{rem:abel}
If $A$ is abelian, then $[A,L_n]=\{1\}$ and so the recursion simplifies to 
$L_{n+1}=\langle  [A, \gamma_{n}(C)], [L_{n},C]\rangle$. On the other hand, 
if $C$ is abelian, then $\gamma_{n}(C)=\{1\}$ for $n\ge 2$ and so
$L_{n+1}=\langle   [A,L_{n}], [L_{n},C]\rangle$ for $n\ge 2$.  Finally, 
if both $A$ and $C$ are abelian, then 
$L_{n+1}=[L_{n},C]$ for $n\ge 2$.
\end{remark}

\begin{example}
\label{ex:heisenberg}
Let $G=F_2/\gamma_3(F_2)$ be the $2$-step free nilpotent group 
of rank $2$. This group can be realized as a split extension of the form 
$\mathbb{Z}^2\rtimes_{\varphi} \mathbb{Z}$, with monodromy given by the matrix 
$\big(\begin{smallmatrix}  1 & 1\\0& 1  \end{smallmatrix}\big)$.
It is readily seen that $L_2=\mathbb{Z}$ and $L_3=0$. 
\end{example}

The next two results are proved in \cite[Lemma 3.3]{GP}. For completeness, 
and since we will need the basic outline of the arguments later on, we 
provide self-contained proofs.

\begin{lemma}[\cite{GP}]
\label{lem:gp-normal} 
The groups $L_n$ are normal subgroups of $A$.
\end{lemma}

\begin{proof}
We prove the claim by induction on $n$, with the base case $L_1=A$ 
tautologically true. So assume $L_n \triangleleft A$.  As noted 
in \S\ref{subsec:comms}, the commutator of two normal subgroups 
is again a normal subgroup; thus, $[A,L_n]  \triangleleft A$. 
Furthermore, by Lemma \ref{lem:normal}, $[A,\gamma_{n}(C)]  \triangleleft A$. 
Finally, if $x\in A$, then $x[L_n,C] x^{-1}$ is contained in $[L_n,C] \cdot [A,L_n]$, 
by formula \eqref{eq:conj-2}, and thus  $x[L_n,C] x^{-1}\subseteq L_{n+1}$. 
Therefore, in view of definition \eqref{eq:subgroup}, we have that 
$L_{n+1}\triangleleft A$, and the induction is complete.
\end{proof}

\begin{lemma}[\cite{GP}]
\label{lem:gp-series} 
The subgroups $\{L_n\}_{n\ge 1}$ form a descending series for $A$. 
\end{lemma}

\begin{proof}
We need to show that, for each $n\ge 1$, we have an inclusion 
$L_{n+1} \subseteq L_{n}$, or, equivalently,  that $[A, L_n]$, 
$[A,\gamma_n(C)]$, and $[L_n,C]$ are all included in $L_n$. 
We consider the three cases in turn.

\begin{enumerate}[(a)]
\item \label{dd1}
$[A, L_n]\subseteq L_n$: follows at once from Lemma \ref{lem:gp-normal} .

\item  \label{dd2}
$[A,\gamma_n(C)]\subseteq L_n$: we prove this by induction on $n$, 
with the base case, $[A,C]\subseteq A$, being obvious. We have 
$[[A,C],\gamma_n(C)]\subseteq [A,\gamma_n(C)] \subseteq L_{n+1}$, 
by \eqref{eq:subgroup} and 
$[[A,\gamma_n(C)], C]\subseteq [L_n,C] \subseteq L_{n+1}$, 
by the induction hypothesis and \eqref{eq:subgroup}. 
Thus,  $[A,\gamma_{n+1}(C)]=[A,[C,\gamma_n(C)]]$ is contained in $L_{n+1}$, 
by Lemma \ref{lem:3subgroups}, and the induction is complete.

\item  \label{dd3}
$[L_n,C] \subseteq L_n$: we prove this again by induction on $n$, with 
the base case $n=1$ being obvious. For the induction step, 
there are three sub-cases to consider. 
\begin{enumerate}[(i)]
\item \label{la1}
Clearly $[[A,C], L_n]\subseteq [A,L_n]$, while  
$[A,[L_n, C]]\subseteq  [A,L_n]$ by the induction hypothesis. 
Therefore, $[[A,L_n], C]\subseteq [A,L_n] \subseteq  L_{n+1}$, 
by Lemma \ref{lem:3subgroups} and \eqref{eq:subgroup}.
\item \label{la2}
We have $[[A,C],\gamma_n(C)]\subseteq [A,\gamma_n(C)]\subseteq L_{n+1}$, 
by \eqref{eq:subgroup}, 
and $[A,[C,\gamma_n(C)]]=[A,\gamma_{n+1}(C)]\subseteq L_{n+1}$,  
by case \ref{dd2}.  Therefore, 
$[[A,\gamma_n(C)], C]\subseteq L_{n+1}$, by Lemma \ref{lem:3subgroups}.
\item \label{la3}
Finally, $[[L_n,C],C]\subseteq [L_n,C] \subseteq L_{n+1}$, 
by the induction hypothesis and \eqref{eq:subgroup}.
\end{enumerate}
This shows  that $[L_{n+1} ,C]\subseteq L_{n+1}$, thereby completing 
the induction for case \ref{dd3}.
\end{enumerate}
This ends the proof.
\end{proof}

\subsection{The series $L$ is an $N$-series}
\label{subsection:l-is-n}
Our next objective is to show that the Guaschi--Pereiro series is, 
in fact, an $N$-series. We start with a preparatory lemma, which 
will be of much use later on.

\begin{lemma}
\label{lem:kngn} 
$[L_{n},\gamma_{m}(C)]\subseteq L_{n+m}$, for all $n,m \ge 1$.
\end{lemma}

\begin{proof}
We prove the  claim by induction on $m$. The base case $m=1$, 
which amounts to $[L_n, C]\subseteq L_{n+1}$ for all $n\ge 1$, 
follows directly from \eqref{eq:subgroup}. Assume now that 
$[L_{n},\gamma_{m}(C)]\subseteq L_{n+m}$, for all $n\ge 1$. Then
$[[L_{n},C],\gamma_{m}(C)] 
\subseteq [L_{n+1},\gamma_{m}(C] \subseteq L_{n+m+1}$, 
and likewise $ [[L_{n},\gamma_{m}(C)],C] 
\subseteq [L_{n+m},C] \subseteq  L_{n+m+1}$. 
Therefore, by Lemma \ref{lem:3subgroups},  
$[L_{n},\gamma_{m+1}(C)]=[L_{n},[C,\gamma_{m}(C)]]$ is 
contained in $L_{n+m+1}$, and so  
the induction step is complete.
\end{proof}

\begin{theorem}
\label{thm:gp-nseries}
The subgroups $\{L_n\}_{n\ge 1}$ form an $N$-series for $A$. 
\end{theorem}

\begin{proof}
From Lemma \ref{lem:gp-series}, we know that $L=\{L_n\}_{n\ge 1}$ is 
a descending series of (normal) subgroups of $A$. Thus, 
we only need to show that $[L_n, L_m]\subseteq L_{n+m}$ for 
all $n,m\ge 1$. 

We prove this claim by induction on $m$. The base case $m=1$, 
which amounts to $[L_n, A]\subseteq L_{n+1}$ for all $n\ge 1$, is built in 
definition \eqref{eq:subgroup}. Assume now that $[L_n, L_m]\subseteq L_{n+m}$ 
for all $n\ge 1$, or, equivalently, that the subgroups 
$[L_n, [A,L_{m-1}]]$, $[L_n, [A,\gamma_{m-1}(C)]]$, and 
$[L_n, [L_{m-1},C]]$ are contained in $L_{n+m}$, 
for all $n\ge 1$. For the induction step, we use 
repeatedly Lemmas \ref{lem:3subgroups} and \ref{lem:kngn}, 
as well as \eqref{eq:subgroup}.
There are three cases to consider.

\begin{enumerate}[itemsep=4pt, topsep=3pt, partopsep=3pt, label=(\alph*)]
\item \label{az1}
$[[L_{n}, A],L_{m}] \subseteq  [L_{n+1},L_m]  \subseteq L_{n+1+m}$ 
and $ [[L_{n}, L_m],A]    \subseteq   [L_{n+m},A]\subseteq L_{n+1+m}$. 
Therefore, $[L_n,[A,L_{m}]]\subseteq L_{n+1+m}$. 
\item \label{az2}
$[[L_n, A], \gamma_{m}(C)] \subseteq [L_{n+1},\gamma_{m}(C)] 
\subseteq L_{n+1+m}$ and $[[L_n, \gamma_{m}(C)], A]
\subseteq [L_{n+m},A] \subseteq L_{n+1+m}$. 
Therefore, $[L_n,[A, \gamma_{m}(C)]]  \subseteq L_{n+1+m}$.
\item \label{az3}
$[[L_{n}, C],L_{m}] \subseteq   [L_{n+1},L_{m}] \subseteq L_{n+1+m}$
and $ [[L_{n}, L_m],C]  \subseteq   [L_{n+m},C]   \subseteq L_{n+1+m}$. 
Therefore, $[L_{n},[L_{m}, C]] \subseteq L_{n+1+m}$.
\end{enumerate}
This shows that $[L_n,L_{m+1}]\subseteq L_{n+m+1}$ for all $n\ge 1$, 
thereby completing the induction.
\end{proof}

We conclude this subsection with one more lemma.

\begin{lemma}
\label{lem:akb} 
The inclusions $\gamma_n(A) \subseteq L_n\subseteq \gamma_n(B)$ 
hold for all $n\ge 1$.
\end{lemma}

\begin{proof}
The first claim follows at once from Lemma \ref{lem:fastest-n} and 
Theorem \ref{thm:gp-nseries}.

The second claim is proved again 
by induction on $n$, with the base case $n=1$ being obvious. 
Assuming $L_{n}\subseteq \gamma_{n}(B)$, we have 
that all three subgroups generating $L_{n+1}$ in \eqref{eq:subgroup} 
are included in $[B,\gamma_{n}(B)]$, whence 
$L_{n+1}\subseteq \gamma_{n+1}(B)$.
\end{proof}

\subsection{The lower central series of a split extension}
\label{subsec:split-gp}
A recent result of Guaschi and Pereiro (\cite[Theorem 1.1]{GP})
expresses the lower central series of a semidirect product 
in terms of the lower central series of the factors and 
the extension data. For completeness, and since we will 
use this approach in both the rational and modular contexts, 
we provide a self-contained proof of this theorem. 
The proof given here is by and large modeled on the original 
one, yet it is more condensed.

\begin{theorem}[\cite{GP}]
\label{thm:gu-pe}
Let $B=A\rtimes_{\varphi} C$ be a split extension of groups. 
For each $n\ge 1$, the following hold.
\begin{enumerate}
\item \label{gp1}
The homomorphism $\varphi\colon C\to \Aut(A)$ 
restricts to a homomorphism  $\varphi\colon \gamma_n(C)\to \Aut(L_n)$.
\item  \label{gp2}
$\gamma_n(B)=L_n\rtimes_{\varphi} \gamma_n(C)$, where 
$\varphi$ is the monodromy action from part \eqref{gp1}.
\end{enumerate}
\end{theorem}

\begin{proof}
We prove claim \eqref{gp1} by induction on $n$, the case $n=1$ being 
tautological.  So assume the map $\varphi\colon C\to \Aut(A)$ restricts 
to a map $\varphi\colon \gamma_{n-1}(C)\to \Aut(L_{n-1})$.  Let 
$g$ be in $\gamma_{n}(C)$, and thus in $\gamma_{n-1}(C)$, too. 
There are three cases to consider. 

\begin{enumerate}[(a)]
\item \label{aaa}
Since $\varphi(g)$ leaves the subgroup $L_{n-1}$ invariant, 
formula \eqref{eq:act-1} implies that $\varphi(g)$ also 
leaves $[A,L_{n-1}]$ invariant.
\item \label{bbb}
Since $gc\in \gamma_{n-1}(C)$ for $c\in \gamma_{n-1}(C)$, 
formula \eqref{eq:act-2} and Lemma \ref{lem:normal} imply 
that $\varphi(g)$ leaves $[A,\gamma_{n-1}(C)]$ invariant.
\item \label{ccc}
By Lemma \ref{lem:normal}, $[L_{n-1},C]$ is a normal subgroup of $A$. 
Formula \eqref{eq:act-2}  now implies that $\varphi(g)$ leaves 
$[L_{n-1},C]$ invariant.
\end{enumerate}

Given the way the subgroup $L_n$ was defined in \eqref{eq:subgroup}, it 
follows that $\varphi(g)$ leaves it invariant. 
Since the map $\varphi(g)\colon A\to A$ is injective, its restriction 
to $L_n$ is also injective. To show that the map $\varphi(g)\colon L_n\to L_n$ 
is surjective, we go through the above three cases one more time. In case 
\ref{aaa}, the surjectivity of the maps  $\varphi(g)\colon A\to A$  
and $\varphi(g)\colon L_{n-1}\to L_{n-1}$ yield the claim. 
In cases \ref{bbb} and \ref{ccc} the claim follows from \eqref{eq:act-2}, by 
observing that $[c,a]=\varphi(g)(x)$, where 
$x=[g^{-1}c,a] \cdot  {}^a[g^{-1},a^{-1}]$.

Claim \eqref{gp2} is also proved by induction on $n$, 
with the case $n=1$ being tautological. Assume that 
$\gamma_{n-1}(B)=L_{n-1}\rtimes_{\varphi} \gamma_{n-1}(C)$. 
To show that $L_{n}\rtimes_{\varphi} \gamma_{n}(C)
\subseteq \gamma_{n}(B)$, observe that 
$\gamma_n(C)\subseteq \gamma_n(B)$,  
and also $L_n\subseteq \gamma_n(B)$, by Lemma \ref{lem:akb}. 
For the reverse inclusion, the induction hypothesis and 
definition \eqref{eq:subgroup} show that
\begin{equation}
\label{eq:gamma-n-b}
\gamma_n(B)=[B,\gamma_{n-1}(B)]\subseteq 
[A\rtimes_{\varphi}  C,L_{n-1}\rtimes_{\varphi} \gamma_{n-1}(C)]
\subseteq L_n \rtimes_{\varphi} \gamma_n(C) \,.
\end{equation}
This completes the proof.
\end{proof}

\section{The associated graded Lie algebra of a split extension}
\label{sect:gr-split}

\subsection{Split extensions of Lie algebras}
\label{subsec:split-lie}

Before proceeding, let us review some notions from the realm  
of Lie algebras. Given a graded Lie algebra $\mathfrak {g}$, define 
the Lie algebra of degree-$0$ derivations, $\Der(\mathfrak {g})$, 
as the additive group of degree-preserving endomorphisms 
$\delta\colon \mathfrak {g} \to \mathfrak {g}$ with the property that 
$\delta [x,y]= [\delta x, y]+ [x, \delta y]$ for all $x,y\in \mathfrak {g}$,  
and with Lie bracket given by 
$[\delta, \delta']= \delta \circ\delta'- \delta' \circ \delta$. 
Now let 
\begin{equation}
\label{eq:lie-exact}
\begin{tikzcd}[column sep=24pt]
0\ar[r] &  \mathfrak {n} \ar[r, "\iota"]
& \mathfrak {g} \ar[r, "\pi"] & {\mathfrak{h}} \ar[r] \arrow[bend right=55,"\tau",swap]{l} & 0
\end{tikzcd}
\end{equation}
be a split exact sequence of graded Lie algebras. Then $\mathfrak {g}$ can be 
identified with the semidirect product $\mathfrak {n}\rtimes_{\theta} {\mathfrak{h}}$, where 
the monodromy $\theta\colon {\mathfrak{h}} \to \Der(\mathfrak {n})$ is the Lie algebra map 
defined by 
\begin{equation}
\label{eq:lie-mono}
\iota(\theta(y)(x))=[\tau(y),\iota(x)]
\end{equation} 
for $x\in \mathfrak {n}$ and $y\in {\mathfrak{h}}$. That is, $\mathfrak {g}=\mathfrak {n}\times {\mathfrak{h}}$ as an abelian group, 
with Lie bracket given by $[(x_1,y_1), (x_2,y_2)] = 
([x_1,x_2]+\theta(y_1)(x_2)-\theta(y_2)(x_1), [y_1,y_2])$.

\subsection{The associated graded Lie algebra of a split extension}
\label{subsec:n-series-split}

Returning to our previous setup, consider a 
split extension of groups, $B=A\rtimes_{\varphi} C$, and recall that 
such an extension defines an $N$-series, $L=\{L_n\}_{n\ge 1}$, for $A$. 
Let $\gr^L(A)$ be the graded Lie algebra associated to this 
$N$-series according to formula \eqref{eq:grPhi}.  
As noted in Lemma \ref{lem:akb}, the inclusion map 
$\alpha\colon A\to B$ sends $L_n$ to $\gamma_n(B)$. 
We thus have an induced morphism of associated graded 
Lie algebras, $\gr^L(\alpha)\colon \gr^{L}(A) \to \gr(B)$. 

\begin{theorem}
\label{thm:gp-gr}
Let $\begin{tikzcd}[column sep=16pt]
\! 1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\!\end{tikzcd}$ 
be a split exact sequence of groups, with monodromy 
$\varphi\colon C\to \Aut(A)$, and let $L$ 
be the $N$-series for $A$ defined in \eqref{eq:subgroup}.
There is then an induced split exact sequence of graded Lie algebras, 
\begin{equation}
\label{eq:split-gr}
\begin{tikzcd}[column sep=16pt]
\!\!0\ar[r] &  \gr^{L}(A) \ar[rr, "\gr^{L}(\alpha)"] &&
\gr(B) \ar[rr, "\gr(\beta)"] && \gr(C) \ar[r] & 0\, .
\end{tikzcd}
\end{equation}
Consequently, $\gr(B) \cong 
\gr^{L}(A) \rtimes_{\bar\varphi} \gr(C)$, where the monodromy 
$\bar\varphi\colon \gr(C) \to \Der(\gr^L(A))$ is the 
map of Lie algebras induced by $\varphi$.
\end{theorem} 

\begin{proof}
By Theorem \ref{thm:gu-pe}, for each $n\ge 1$, there is a 
split exact sequence,
\begin{equation}
\label{eq:split-ell-n}
\begin{tikzcd}[column sep=20pt]
1\ar[r] & L_n \ar[r, "\alpha"]
& \gamma_n(B) \ar[r, "\beta"] & \gamma_n(C) \ar[r] & 1 \, .
\end{tikzcd} 
\end{equation}
For each $n\ge 1$, the Snake Lemma yields split exact sequences 
of abelian groups, 
\begin{equation}
\label{eq:snake}
\begin{tikzcd}[column sep=16pt]
\!\!0\ar[r] &  L_n/L_{n+1} \ar[rr, "\gr^{L}_n(\alpha)"] &&
\gamma_n(B)/\gamma_{n+1}(B) \ar[rr, "\gr_n(\beta)"] 
&& \gamma_n(C)/\gamma_{n+1}(C) \ar[r] & 0 \, .
\end{tikzcd}
\end{equation}
In turn, these sequences assemble in the exact 
sequence \eqref{eq:split-gr} of graded Lie algebras. 

If $\sigma\colon C\to B$ is a splitting for the map 
$\beta$, the above construction induces a splitting 
$\gr(\sigma)\colon \gr(C)\to \gr(B)$ for $\gr(\beta)$. 
The homomorphism $\varphi\colon C \to \Aut(A)$ defined by 
\eqref{eq:action} in terms of the splitting $\sigma$ and 
conjugation in $B$ induces a morphism of graded Lie algebras, 
$\bar\varphi\colon \gr(C) \to \Der(\gr^{L}(A))$,  defined by 
\eqref{eq:lie-mono} in terms of the splitting $\gr(\sigma)$ and 
the Lie bracket in $\gr(B)$. This realizes 
the associated graded Lie algebra of $B$ as a split extension  
of $\gr(C)$ with $\gr^L(A)$ with monodromy $\bar\varphi$, as claimed.
\end{proof}

\subsection{Discussion and examples}
\label{subsec:discuss}
In general, we cannot replace $\gr^{L}(A)$ by $\gr(A)$ in Theorem \ref{thm:gp-gr}. 
Indeed, the natural map of graded Lie algebras, $\gr(A) \to \gr^{L}(A)$,   
need not be an isomorphism, due to the fact that the inclusions 
$\gamma_n(A) \hookrightarrow L_n$ need not be surjective. We illustrate this 
phenomenon with a few examples.

\begin{example}
\label{ex:klein}
Let $K=\langle a,t \mid tat^{-1} =a^{-1}\rangle$ be the fundamental group of the 
Klein bottle. Then $K$ is a split extension of the form $A\rtimes_{\varphi} C$, 
where $C=\langle t\rangle$ acts by inversion on $A=\langle a\rangle$.  
It is readily seen that $L_n=\langle a^{2^{n-1}}\rangle$ for $n\ge 1$, and 
thus $\gr^L_n(A)=\mathbb{Z}_2$ for $n\ge 1$, although of course $\gamma_n(A)=\{1\}$ 
and $\gr_n(A)=0$ for $n>1$.  It also follows from Theorem \ref{thm:gu-pe} that 
$\gamma_n(K)=L_n$ for $n>1$, and thus $\gamma_{\omega}(K)=\{1\}$. 
\end{example}

\begin{example}
\label{ex:knot}
Let $G=\pi_1(X)$ be the fundamental group of a knot complement, 
$X=S^3 \setminus K$.  For simplicity, we assume the knot is fibered 
and non-trivial, so that $X$ is the mapping torus of a homeomorphism 
$h\colon S\to S$, where $S$ is a punctured orientable surface of 
genus $g>0$.  Then $G=F\rtimes_{\varphi} \mathbb{Z}$, where $F=\pi_1(S)$  
is a free group of rank $2g$ and the monodromy 
$\varphi\colon \mathbb{Z}\to \Aut(F)$ sends $1$ to the automorphism 
$h_*\colon \pi_1(S)\to \pi_1(S)$.  The projection map $\beta\colon G\to \mathbb{Z}$ 
induces an isomorphism $\beta_{\ab}\colon G_{\ab}\isom \mathbb{Z}$, which extends  
to an isomorphism $\gr(\beta) \colon \gr(G)\isom \gr(\mathbb{Z})$, concentrated in 
degree $1$.  Clearly, $\gr(F)=\Lie(\mathbb{Z}^{2g})$ yet $\gr^{L}(F)=0$, 
showing how far apart these two Lie algebras are in this case.  
\end{example}

\begin{example}
\label{ex:fp-poison}
The Formanek--Procesi ``poison group" is a split extension of the form 
$G=A \rtimes_{\varphi} C$, with factors $A=F_3=\langle a_1,a_2,a_3\rangle$  
and $C=F_2=\langle c_1,c_2\rangle$, and with monodromy action given by 
$\varphi(c_i)(a_j)=a_j$ and $\varphi(c_i)(a_3)=a_3a_i$, for $1\le i,j\le 2$. 
Its notoriety comes from the fact that $G$ (and thus $\Aut(F_n)$ for $n\ge 3$, 
or any other group in which this poison group embeds) does not admit a 
finite-dimensional linear representation. In \cite[Proposition 7.1]{CCP}, Cohen, 
Cohen, and Prassidis analyzed  
the associated graded Lie algebra of $G$, and concluded that the kernel of the 
projection map $\gr(G)\to \gr(C)$ is $\Lie(\mathbb{Z})=\mathbb{Z}$, generated by $a_3$. By 
Theorem \ref{thm:gp-gr}, though, this kernel is the Lie algebra $\gr^{L}(A)$, 
which indeed has degree $1$ piece equal to $\mathbb{Z}=\langle a_3\rangle$, but 
also has degree $2$ piece equal to $\mathbb{Z}^2=\langle [a_1,a_3],[a_2,a_3]\rangle$, 
and so on. In fact, it follows from \cite{Sev,Met-Pap} that 
$\gr^{L}_{n}(A) \cong \Lie_{n-1}(\mathbb{Z}^2)$ for $n\ge 2$, although 
of course $\gr(A)=\Lie(\mathbb{Z}^3)$. 
\end{example}

Nevertheless, as we shall see in the next section, there is a noteworthy class of split 
extensions of the form $A\rtimes C$ for which all the maps 
$\gamma_n(A) \hookrightarrow L_n$ are isomorphisms,  and thus $\gr(A) \cong \gr^{L}(A)$.

\section{Trivial action on abelianization}
\label{sect:ext}

\subsection{$N$-series for almost direct products}
\label{subsec:almost-direct}
Once again, let $B=A\rtimes_{\varphi} C$ be a split extension of groups, 
with monodromy $\varphi\colon C\to \Aut(A)$. Following \cite{FR}, 
we say that $B$  is an {\em almost direct product}\/ of $C$ and $A$ if 
$C$ acts trivially on the abelianization $A_{\ab}=H_1(A,\mathbb{Z})$. In other 
words, the monodromy of the extension factors through a map 
$\varphi\colon C\to \mathcal{T}(A)$, where $\mathcal{T}(A)\coloneq 
\ker \big(\!\Aut(A) \to \Aut(A_{\ab})\big)$ is the Torelli group of $A$. 

On elements, the above condition says that 
$\varphi(c)(a) \cdot a^{-1}  \in A'$, for all $c\in C$ and $a\in A$. 
If we view $C$ as a subgroup of $G$ via the splitting $\sigma\colon C\to B$, 
so that $\varphi(c)(a)\cdot a^{-1}=[c,a]$, the condition most succinctly reads as 
\begin{equation}
\label{eq:ac-gamma}
[A,C]\subseteq \gamma_2(A)\,.
\end{equation}

As noted in \cite[Proposition 6.3]{BGG11}, the property of a split extension 
$B=A\rtimes_{\varphi} C$ being an almost direct product does not depend 
on the choice of splitting. That is, if $\sigma'\colon C\to B$ 
is another splitting of the projection $\beta \colon B\to C$, and if 
$\varphi'\colon C\to \Aut(A)$ is the corresponding monodromy action, 
then the split extension $B=A\rtimes_{\varphi'} C$ is again an almost 
direct product. 

Here is the key result of this section.

\begin{theorem}
\label{thm:fr-lem}
Let $B=A\rtimes_{\varphi} C$ be an almost direct product of groups, and 
let $L=\{L_n\}_{n\ge 1}$ be the corresponding $N$-series for $A$. 
Then $L_n=\gamma_n(A)$ for each $n\ge 1$.
\end{theorem}

\begin{proof}
By Lemma \ref{lem:akb}, we have that $\gamma_n(A)\subseteq L_n$, 
for all $n\ge 1$. We prove the reverse inclusion by induction on $n$, 
with the base case $n=1$ being obvious. So assume 
$L_{r}\subseteq \gamma_{r}(A)$ for $r\le n$; in view 
of \eqref{eq:subgroup}, for $r=n$ this condition is equivalent to 
\begin{equation}
\label{eq:induct-fr}
[\gamma_{n-1}(A),C]\subseteq \gamma_{n}(A), \quad
[A,\gamma_{n-1}(C)]\subseteq \gamma_{n}(A)\, .
\end{equation} 

By \eqref{eq:induct-fr},  $[[A,\gamma_{n-1}(C)],A]$ is included in 
$[\gamma_{n}(A),A] =\gamma_{n+1}(A)$. Thus, by Lemma \ref{lem:3subgroups}, 
$[[A,A],\gamma_{n-1}(C)]$ is also included in $\gamma_{n+1}(A)$; that is, 
\begin{equation}
\label{eq:g2}
[\gamma_2(A),\gamma_{n-1}(C)] \subseteq\gamma_{n+1}(A)\, .
\end{equation}  

For the induction step, it is enough  to show that all three subgroups 
from \eqref{eq:subgroup} that generate $L_{n+1}$ are included 
in $\gamma_{n+1}(A)$; we do this next, keeping in mind that, 
by the induction hypothesis, $L_{n}=\gamma_{n}(A)$. 
\begin{enumerate}[(a)]
\item \label{a}
$[A,\gamma_{n}(A)] \subseteq \gamma_{n+1}(A)$. This follows at once from 
the fact that $\gamma(A)$ is an $N$-series.
\\[-8pt]
\item  \label{b}
$[\gamma_{n}(A),C]\subseteq \gamma_{n+1}(A)$.  We have 
\begin{align*}
&[[A,C],\gamma_{n-1}(A)] 
\underset{ \eqref{eq:ac-gamma}}\subseteq [\gamma_2(A),\gamma_{n-1}(A)] 
= \gamma_{n+1}(A),
\\
&[[\gamma_{n-1}(A),C],A] 
\underset{ \eqref{eq:induct-fr}}\subseteq [\gamma_{n}(A),A] 
= \gamma_{n+1}(A).
\end{align*}
Hence, by Lemma \ref{lem:3subgroups}, $[\gamma_{n}(A),C] =[[\gamma_{n-1}(A),A],C]$ 
is included in $\gamma_{n+1}(A)$.
\\[-8pt]
\item  \label{c}
$[A,\gamma_{n}(C)]\subseteq \gamma_{n+1}(A)$.  
 We have
\begin{align*}
&[[A,C],\gamma_{n-1}(C)] 
\underset{ \eqref{eq:ac-gamma}}\subseteq [\gamma_2(A),\gamma_{n-1}(C)] 
\underset{\eqref{eq:g2}}\subseteqq \gamma_{n+1}(A),
\\
&[[A,\gamma_{n-1}(C)],C] 
\underset{ \eqref{eq:induct-fr}}\subseteq [\gamma_{n}(A),C] 
\underset{\ref{b}}\subseteq\gamma_{n+1}(A). 
\end{align*}
Hence,  by Lemma \ref{lem:3subgroups}, 
$[A,\gamma_{n}(C)]=[A,[\gamma_{n-1}(C),C]]$ is 
included in $\gamma_{n+1}(A)$.
\end{enumerate}
This verifies the induction step, and thus completes the proof.
\end{proof}

\subsection{Applications}
\label{subsec:almost-direct-fr}
We now derive several applications of Theorem \ref{thm:fr-lem}, 
based on the results from the previous section. We start by 
recovering a technical result of Falk and Randell 
(\cite[Lemma 3.4]{FR}; the case $m=1$ is \cite[Lemma 3.3]{FR}).

\begin{corollary}[\cite{FR}]
\label{cor:fr-lemma}
Let $B=A\rtimes C$ be an almost direct product.  Then 
$[\gamma_{n}(A),\gamma_{m}(C)]\subseteq \gamma_{n+m}(A)$,  
for all $n,m \ge 1$.
\end{corollary}

\begin{proof}
By Theorem \ref{thm:fr-lem}, $\gamma_n(A)=L_n$. The 
claim now follows from Lemma \ref{lem:kngn}.
\end{proof}

Next, we recover a well-known result of Falk and Randell \cite[p.~85]{FR}. 

\begin{corollary}[\cite{FR}]
\label{cor:split-exact-lcs}
Let $B=A\rtimes_{\varphi} C$ be an almost direct product. Then 
\begin{enumerate}
\item \label{fr1}
$\gamma_n(B)=\gamma_n(A)\rtimes_{\varphi} \gamma_n(C)$ 
for all $n\ge 1$.
\item \label{fr2}
The corresponding  split exact sequence,
$\!\!\begin{tikzcd}[column sep=16pt]
1\ar[r] & A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1 ,
\end{tikzcd}\!$ restricts to split exact sequences 
$\!\!\begin{tikzcd}[column sep=16pt]
1\ar[r] &  \gamma_n(A)\ar[r, "\alpha"]
& \gamma_n(B) \ar[r, "\beta"] & \gamma_n(C) \ar[r] & 1
\end{tikzcd}\!\!$ for all $n\ge 1$.
\end{enumerate}
\end{corollary}

\begin{proof}
Follows at once from Theorems \ref{thm:gu-pe} and \ref{thm:fr-lem}.
\end{proof}

The following corollary extends another result of Falk and 
Randell (\cite[Theorem 2.6]{FR88}) to a more general setting.

\begin{corollary}[\cite{FR88}]
\label{cor:res-nilp}
Suppose $B=A\rtimes_{\varphi} C$ is an almost direct product of two 
residually nilpotent groups. Then $B$ is also residually nilpotent.
\end{corollary}

\begin{proof}
Let $\!\!\begin{tikzcd}[column sep=18pt]
1\ar[r] & A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1 ,
\end{tikzcd}\!$
be the corresponding split exact sequence. 
By Lemma \ref{lem:res-nilp}, we have that 
$\gamma_{\omega}(A)=\{1\}$  and 
$\gamma_{\omega}(C)=\{1\}$. 
Let $g\in \gamma_{\omega}(B)$. Then 
$\beta(g)\in \gamma_{\omega}(C)$, and so $\beta(g)=1$.
Thus, there is an element $a\in A$ such that $\alpha(a)=g$. 

Now fix an index $n\ge 1$. Then $g\in \gamma_n(B)$, and since 
$\beta(g)=1$, Corollary \ref{cor:split-exact-lcs} insures the existence 
of an element $a_n\in \gamma_n(A)$ such that $\alpha(a_n)=g$. 
But the map $\alpha\colon A\to B$ is injective, and so we 
must have $a=a_n$, whence  $a\in \gamma_n(A)$. 
It follows that $a\in \gamma_{\omega}(A)$, and thus $a=1$. 
Therefore, $g=1$, and we are done.
\end{proof}

Next, we recover another well-known result of Falk and Randell 
(\cite[Theorem 3.1]{FR}), in a slightly stronger form.

\begin{corollary}[\cite{FR}]
\label{cor:falk-ran}
Let $\begin{tikzcd}[column sep=16pt]
\! 1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\!\end{tikzcd}$ 
be a split exact sequence of groups.  Suppose the resulting action, 
$\varphi\colon C\to \Aut(A_{\ab})$, is trivial. Then the sequence 
of induced maps, 
\begin{equation}
\label{eq:split-lie}
\begin{tikzcd}[column sep=28pt]
\!\!0\ar[r] &  \gr(A) \ar[r, "\gr(\alpha)"] &
\gr(B) \ar[r, "\gr(\beta)"] & \gr(C) \ar[r] & 0 ,
\end{tikzcd}
\end{equation}
is a split exact sequence of graded Lie algebras. Thus, 
$\gr(B) \cong \gr(A)\rtimes_{\bar\varphi} \gr(C)$, where 
the monodromy $\bar\varphi\colon \gr(C) \to \Der(\gr(A))$ 
is the map of Lie algebras induced by $\varphi$.
\end{corollary} 

\begin{proof}
By Theorem \ref{thm:fr-lem}, we have that $L_n=\gamma_n(A)$ for all $n\ge 1$, 
and so $\gr^L(A)=\gr(A)$. Both claims now follow from Theorem \ref{thm:gp-gr}.
\end{proof}

In \cite[Proposition 2]{BB09}, Bardakov and Bellingeri obtain as 
an application of Corollary \ref{cor:split-exact-lcs} the 
following result. 
 
\begin{corollary}[\cite{BB09}]
\label{cor:bb}
Suppose $B=A\rtimes_{\varphi} C$ is an almost direct product. Then 
$\!\sqrt{\gamma_n(B)}=\!\sqrt{\gamma_n(A)}\rtimes_{\varphi} \!\sqrt{\gamma_n(C)}$.
\end{corollary}

We will strengthen this result  in Theorem \ref{theorem:fr-rational}, 
where we will show that the hypothesis that $C$ acts trivially on 
$A_{\ab}$ can be replaced by the much weaker hypothesis that $C$ 
acts trivially on $A_{\abf}$.

\section{The rational lower central series of a split extension}
\label{sect:rat-split}

\subsection{A rational Guaschi--Pereiro series}
\label{subsec:rat-gp}
In this section we study  the rational lower series of a 
a split extension of groups, $B=A\rtimes_{\varphi} C$. 
Let $L=\{L_n\}_{n\ge 1}$ be the sequence of subgroups 
of $A$ defined by the recursion formula \eqref{eq:subgroup}. 
We will focus on the sequence $\{\!\sqrt{L_n}\}_{n\ge 1}$, 
where $\!\sqrt{L_n}=\!\sqrt[A]{L_n}$ is the isolator of $L_n$ in $A$. 
By definition, 
\begin{equation}
\label{eq:sqrt-ln}
\!\sqrt{L_{n+1}} = 
\sqrt{ \langle [A,L_{n}] , [A,\gamma_{n}(C)], [L_{n},C]\rangle } \, .
\end{equation}
Thus, every element in $\!\sqrt{L_{n+1}}$ admits a non-trivial power which 
can be expressed as a word in elements  from the subgroups 
$[A,L_{n}]$, $[A,\gamma_{n}(C)]$, and $[L_{n},C]$. 
We start with a crucial fact about this series.

\begin{lemma}
\label{lem:sqrt-kn}
The sequence $\!\sqrt{L}=\{\!\sqrt{L_n}\}_{n\ge 1}$ is 
an $N_0$-series for $A$.
\end{lemma}

\begin{proof}
Follows at once from Proposition \ref{prop:sqrt-N} and 
Theorem \ref{thm:gp-nseries}.
\end{proof}

From Lemma \ref{lem:kngn}, we know that 
$[L_n,\gamma_m(C)]\subseteq L_{n+m}$ for all $n,m\ge 1$. 
The next two lemmas constitute a rational analogue of this result.

\begin{lemma}
\label{lem:srqtL-gammaC}
$[\!\sqrt[A]{L_n},\gamma_m(C)]\subseteq \!\sqrt[A]{L_{m+n}}$ for all $m,n\ge 1$.
\end{lemma}

\begin{proof}
The proof is by induction on $n$. Since $\!\sqrt{L_1}=A$, the base case $n=1$ 
follows at once from Lemma \ref{lem:kngn}. So assume that 
$[\!\sqrt{L_n},\gamma_m(C)]\subseteq \!\sqrt{L_{m+n}}$ for all $m\ge 1$. 
Let $a\in \!\sqrt{L_{n+1}}$, so that $a^k \in L_{n+1}$, for some $k>0$, 
and let $c\in \gamma_m(C)$. Since $a$ belongs to $\!\sqrt{L_{n}}$, too, 
we may apply Lemma \ref{lem:conj-mod} to infer that 
\[
[c,a]^{k} \equiv [c,a^{k}] \mod 
[\!\sqrt{L_{n}}, [\!\sqrt{L_{n}},\gamma_m(C)]].
\] 

By the induction hypothesis, the subgroup on the right is contained in 
$[\!\sqrt{L_{n}},  \!\sqrt{L_{m+n}}]$, which in turn is a 
subgroup of $\!\sqrt{L_{m+2n}} \subseteq \!\sqrt{L_{m+n+1}}$ 
by Lemma \ref{lem:sqrt-kn} and the fact that $n\ge 1$.  On the 
other hand, $[c,a^k]$ belongs to $[L_{n+1},\gamma_m(C)]$, which is a 
subgroup of $L_{m+n+1}$, by Lemma \ref{lem:kngn}. Therefore, 
$[c,a]^{k} \in  \!\sqrt{L_{m+n+1}}$, and so  
$[c,a]\in  \!\sqrt{L_{m+n+1}}$. This shows that 
$[\!\sqrt{L_{n+1}},\gamma_m(C)]\subseteq \!\sqrt{L_{m+n+1}}$, 
thereby completing the induction.
\end{proof}

\begin{lemma}
\label{lem:srqt-comm}
$[\!\sqrt[A]{L_n},\!\sqrt[C]{\gamma_m(C)}]\subseteq \!\sqrt[A]{L_{m+n}}$ 
for all $m,n\ge 1$.
\end{lemma}

\begin{proof}
The proof is by induction on $m$. Since $\!\sqrt[C]{\gamma_m(C)}=C$, 
the base case $m=1$ follows at once from Lemma \ref{lem:srqtL-gammaC}. 
So assume that 
$[\!\sqrt[A]{L_n},\!\sqrt[C]{\gamma_m(C)}]\subseteq \!\sqrt[A]{L_{m+n}}$ 
for all $n\ge 1$. 
Let $c\in \!\sqrt[C]{\gamma_{m+1}(C)}$, so that $c^k \in \gamma_{m+1}(C)$, 
for some $k>0$, and let $a\in \!\sqrt[A]{L_{n}}$. Since $c$ belongs to 
$\!\sqrt[C]{\gamma_m(C)}$, too, we may apply Lemma \ref{lem:conj-mod} 
to deduce that 
\[
[a,c]^{k} \equiv [a,c^{k}] \mod 
\Big[ \Big[\!\sqrt[A]{L_{n}},\!\sqrt[C]{\gamma_m(C)}\Big],\!\sqrt[C]{\gamma_m(C)}\Big] .
\]

Employing the induction hypothesis twice, we see that the subgroup on the right is 
contained in $[[\!\sqrt[A]{L_{m+n}},\!\sqrt[C]{\gamma_m(C)}]]\subseteq \!\sqrt[A]{L_{2m+n}}$, 
which in turn is a subgroup of $\!\sqrt[A]{L_{m+n+1}}$, since $m\ge 1$.  
On the other hand, $[a,c^k]\in [\!\sqrt[A]{L_{n}},\gamma_{m+1}(C)]$, 
which is a subgroup of $\!\sqrt{L_{m+n+1}}$ by Lemma \ref{lem:srqtL-gammaC}. 
Hence, $[a,c]^{k} \in  \!\sqrt{L_{m+n+1}}$, and so  
$[a,c]\in  \!\sqrt{L_{m+n+1}}$. This shows that 
$[\!\sqrt[A]{L_n},\!\sqrt[C]{\gamma_{m+1}(C)}]\subseteq \!\sqrt[A]{L_{m+n+1}}$, 
and the induction is complete.
\end{proof}

\subsection{The rational lower central series of a semidirect product}
\label{subsec:rat-semi}
We are now ready to state and prove the main result of this section---an 
analogue of the Guaschi--Pereiro theorem for the rational lower central series.

\begin{theorem}
\label{thm:gp-rat}
Let $B=A\rtimes_{\varphi} C$ be a semidirect product of groups. 
For each $n\ge 1$, 
\begin{enumerate}
\item \label{gpr1}
The homomorphism $\varphi\colon C\to \Aut(A)$ restricts to homomorphisms 
$\varphi\colon\! \sqrt[C]{\gamma_n(C)}\to \Aut\!\big(\!\sqrt[A]{L_n}\big)$.
\item \label{gpr2}
$\sqrt[B]{\gamma_n(B)}=\sqrt[A]{L_n}\rtimes_{\varphi}\!\sqrt[C]{\gamma_n(C)}$.
\end{enumerate}
\end{theorem}

\begin{proof}
\eqref{gpr1}
From Theorem \ref{thm:gu-pe}, we know that $\varphi$ 
restricts to maps $\varphi\colon \gamma_n(C)\to \Aut(L_n)$, for all $n\ge 1$. 
We prove the first claim by induction on $n$, the case $n=1$ being tautological. 
So assume the map $\varphi\colon C\to \Aut(A)$ restricts to a map 
$\varphi\colon\!\sqrt[C]{\gamma_{n-1}(C)}\to \Aut\!\big(\!\sqrt[A]{L_{n-1}}\big)$.  

Let $g$ be in $\!\sqrt[C]{\gamma_{n}(C)}$, and thus also in $\!\sqrt[C]{\gamma_{n-1}(C)}$.
Let $x \in \!\sqrt[A]{L_{n}}$, so that $x^{\ell}\in L_n$, for some $\ell>0$. 
In view of \eqref{eq:subgroup}, there are three cases to consider. 

\begin{enumerate}[itemsep=4pt, topsep=3pt, partopsep=3pt, label=(\alph*)]
\item \label{a-1}
$x^{\ell} \in [A,L_{n-1}]$.  
Writing $x^{\ell}=[a,a']$ with $a\in A$ and 
$a'\in L_{n-1}$, we have by \eqref{eq:act-1} that 
$\varphi(g)(x^{\ell})\in [A,L_{n-1}]\subseteq L_{n}$.
\item \label{b-1}
$x^{\ell} \in [A,\gamma_{n-1}(C)]$. Writing $x^{\ell}=[c,a]$ 
with $a\in A$ and $c\in \gamma_{n-1}(C)$, we have by 
\eqref{eq:act-2} that 
$\varphi(g)(x^{\ell})=[gc,a]\cdot  {}^a[g,a^{-1}]$.  
By Lemma \ref{lem:normal}, this element belongs to 
$\big[A,\!\sqrt[C]{\gamma_{n-1}(C)}\big]$; thus, 
by Lemma \ref{lem:srqt-comm}, 
$\varphi(g)(x^{\ell})\in \!\sqrt[A]{L_{n}}$.
\item \label{c-1}
$x^{\ell} \in [L_{n-1},C]$. Writing $x^{\ell}=[c,a]$ 
with $a\in L_{n-1}$ and $c\in C$, we have by 
\eqref{eq:act-2} that $\varphi(g)(x^{\ell})\in 
[L_{n-1},C]\subseteq L_{n}$.
\end{enumerate}

Consequently, $\varphi(g)(x)\in \!\sqrt[A]{L_{n}}$, thus showing that 
$\varphi(g)$ leaves $\!\sqrt[A]{L_{n}}$ invariant. 
Since the map $\varphi(g)\colon A\to A$ is injective, its restriction 
to $\!\sqrt[A]{L_{n}}$ is also injective. 
To show that the map $\varphi(g)\colon \!\sqrt[A]{L_{n}} \to \!\sqrt[A]{L_{n}}$ 
is surjective, we go again through the above three cases. 
In case \ref{a-1}, the surjectivity of the maps  $\varphi(g)\colon A\to A$  
and $\varphi(g)\colon L_{n-1}\to L_{n-1}$ yields the claim. 
In cases \ref{b-1} and \ref{c-1}, the claim follows from the  
observation that $\varphi(g)( 
[g^{-1}c,a] \cdot  {}^a[g^{-1},a^{-1}])=[c,a]$.
\\[-4pt]

\eqref{gpr2}
Clearly, $\!\sqrt[C]{\gamma_n(C)}\subseteq \!\sqrt[B]{\gamma_n(B)}$; 
moreover, it follows from Lemma \ref{lem:akb} that $\!\sqrt[A]{L_n} 
\subseteq \!\sqrt[B]{\gamma_n(B)}$. Therefore, 
$\!\sqrt[A]{L_n} \rtimes_{\varphi} \!\sqrt[C]{\gamma_n(C)} 
\subseteq\!\sqrt[B]{\gamma_n(B)}$.

To prove the reverse inclusion, let $b\in \sqrt[B]{\gamma_n(B)}$. 
Since $b\in B=A\rtimes C$, we may write in a unique way $b=ac$, 
with $a\in A$ and $c\in C$. From Theorem \ref{thm:gu-pe}, we 
know that $\gamma_n(B)=L_n\rtimes_{\varphi} \gamma_n(C)$. 
Hence, there is an integer $k>0$ such that $b^k=a'c'$, for 
some (uniquely defined) elements $a'\in L_n$ and 
$c'\in \gamma_n(C)$. Applying Lemma \ref{lem:power} 
with $A_1=A$ and $C_1=C$, we have that 
$(ac)^k \equiv a^k c^k$ modulo $[A,C]$, which is a 
subgroup of $L_2\le A$. Therefore, $a^k c^k \equiv a'c' \bmod L_2$. 
Hence, $c^k=c'\in \gamma_n(C)$, 
and so $c\in \!\sqrt[C]{\gamma_n(C)}$.

Applying once again Lemma \ref{lem:power}, this time with 
$A_1=A$ and $C_1=\gamma_n(C)$, we find that 
$(ac)^k \equiv a^k c^k$ modulo $[A, \!\sqrt[C]{\gamma_n(C)}]$, 
which by Lemma \ref{lem:srqt-comm} is a subgroup of 
$\!\sqrt[A]{L_{n+1}}$. Hence, $a^k\equiv a' \bmod \!\sqrt[A]{L_{n+1}}$; 
since $a'\in L_n$, we conclude that $a\in \!\sqrt[A]{L_n}$.
Therefore, $\sqrt[B]{\gamma_n(B)}$ is included in 
$\!\sqrt[A]{L_n} \rtimes_{\varphi} \!\sqrt[C]{\gamma_n(C)} $, 
and the proof is complete.
\end{proof}

\subsection{Rational associated graded Lie algebras}
\label{subsec:rat-lie-ext}
Recall that $\gr^{\rat}(G)$ denotes the graded Lie algebra associated 
to the $N$-series $\gamma^{\rat}(G)=\!\sqrt{\gamma(G)}$ for $G$. Since 
$\gamma^{\rat}(G)$ is, in fact, an $N_0$-series, each graded piece 
is a torsion-free abelian group. 
Now let $B=A\rtimes_{\varphi} C$, and let $\!\sqrt{L}=\{\!\sqrt{L_n}\}_{n\ge 1}$   
be the sequence of subgroups of $A$ defined by \eqref{eq:subgroup} and 
\eqref{eq:sqrt-ln}.
By Lemma \ref{lem:sqrt-kn}, this is an $N_0$-series for $A$; thus, 
the graded Lie algebra associated to this filtration, $\gr^{\!\sqrt{L}}(A)$, 
also has the property that all its graded pieces are torsion-free. 

Theorem \ref{thm:gp-rat} has the following corollary, whose proof is 
completely analogous to that of Theorem \ref{thm:gp-gr}.

\begin{corollary}
\label{cor:gp-gr-rat}
Let $\begin{tikzcd}[column sep=16pt]
\! 1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\!\end{tikzcd}$ 
be a split exact sequence of groups.
There is then an induced split exact sequence of graded Lie algebras, 
\begin{equation}
\label{eq:split-gr-rat}
\begin{tikzcd}[column sep=30pt]
\!\!0\ar[r] &  \gr^{\!\sqrt{L}}(A) \ar[r, "\gr^{\!\sqrt{L}}(\alpha)"] &
\gr^{\rat}(B) \ar[r, "\gr^{\rat}(\beta)"] & \gr^{\rat}(C) \ar[r] & 0\, .
\end{tikzcd}
\end{equation}
Consequently, $\gr^{\rat}(B) \cong 
\gr^{\!\sqrt{L}}(A) \rtimes_{\bar\varphi} \gr^{\rat}(C)$, where 
$\bar\varphi\colon \gr^{\rat}(C) \to \Der \big(\gr^{\!\sqrt{L}}(A)\big)$ 
is the map of Lie algebras induced by the monodromy 
$\varphi\colon C\to \Aut(A)$ of the split extension.
\end{corollary} 

We illustrate the way this corollary works with a simple example.
\begin{example}
\label{ex:klein-rat}
Let $K=A\rtimes C= \langle a,t \mid tat^{-1} =a^{-1}\rangle$ be the 
Klein bottle group from Example \ref{ex:klein}, with $A=C=\mathbb{Z}$.  
Then $\gr^{\rat}(K)=\gr^{\rat}(C)=\mathbb{Z}$ while $\gr^{\!\sqrt{L}}(A)=0$.
\end{example}

\section{Trivial action on torsion-free abelianization}
\label{sect:rat-fr}

Continuing with the setup and notation from the previous section, 
we now analyze the rational lower central series of groups that 
arise as {\em rational almost-direct products}. By definition, 
these groups are split extensions of the form 
$B=A\rtimes_{\varphi} C$, where $C$ acts trivially on 
the torsion-free abelianization $A_{\abf}=H_1(A,\mathbb{Z})/\Tors$. 
 
This condition is equivalent to $\varphi(c)(a) \cdot a^{-1}  \in \!\sqrt{A'}$, 
for all $c\in C$ and $a\in A$. If we view $C$ as a subgroup of $G$ 
via the splitting $\sigma\colon C\to B$, so that $\varphi(c)(a)\cdot a^{-1}=[c,a]$, 
the condition most succinctly reads as 
\begin{equation}
\label{eq:ac-gamma-sqrt}
[A,C]\subseteq \!\sqrt{\gamma_2(A)}\, .
\end{equation}

\begin{remark}
\label{rem:q-trivial}
Clearly, if $C$ acts trivially on $A_{\abf}$, then it acts trivially on 
$A_{\abf}\otimes \mathbb{Q}$. As we note in \cite{Su-abexact}, the reverse 
implication holds if $A_{\abf}$ is finitely generated, but it is not 
valid in general.
\end{remark}

Arguing as in the proofs of \cite[Proposition 6.3]{BGG11} 
and \cite[Proposition 3.2]{BG}, it is readily seen that 
the property of a split extension being a rational almost 
direct product does not depend on the choice of splitting. 

The next lemma is the key technical result of this section.

\begin{lemma}
\label{lem:act-sqrt}
Assume that $[A,C]\subseteq \!\sqrt{\gamma_2(A)}$.
Then, for all $m, n\ge 1$,
\begin{equation}
\label{eq:gamma-ac}
\Big[ \!\sqrt{\gamma_n(A)}, \gamma_m(C)\Big]\subseteq 
\!\sqrt{\smash[b]\gamma_{m+n}(A)}.
\end{equation}
\end{lemma}

\begin{proof}
We prove the claim by induction on $m$. The case $m=1$, 
which amounts to $[\!\sqrt{\gamma_n(A)},C]\subseteq \!\sqrt{\gamma_{n+1}(A)}$ 
for all $n\ge 1$, is proved by induction on $n$. The base case, 
$[A,C]\subseteq \!\sqrt{\gamma_2(A)}$, is our assumption. 
So assume $[\!\sqrt{\gamma_{n}(A)},C]\subseteq \!\sqrt{\gamma_{n+1}(A)}$. 
Let $a\in  \!\sqrt{\gamma_{n+1}(A)}$, so that $a^k \in \gamma_{n+1}(A)$, for 
some $k>0$, and let $c\in C$. 
Since $a$ belongs to $\!\sqrt{\gamma_{n}(A)}$, too, 
we may apply Lemma \ref{lem:conj-mod} to deduce that 
\[
[c,a]^{k} \equiv [c,a^{k}] \:\:\operatorname{ modulo } 
\Big[\!\sqrt{\gamma_{n}(A)}, \Big[\!\sqrt{\gamma_{n}(A)},C\Big]\Big] ,
\]
a subgroup contained in 
$[\!\sqrt{\gamma_{n}(A)}, \!\sqrt{\gamma_{n+1}(A)}]$, by the 
induction hypothesis. In turn, this is a subgroup of 
$\!\sqrt{\gamma_{2n+1}(A)} \subseteq \!\sqrt{\gamma_{n+2}(A)}$,  
since $\!\sqrt{A}$ is an $N$-series and $n\ge 1$.  
On the other hand, $[c,a^k]\in [\gamma_{n+1}(A),C]$. Using once 
again our assumption and the induction hypothesis, we have 
\begin{align*}
[\gamma_{n}(A), [A,C]] &\subseteq \Big[\gamma_{n}(A), \!\sqrt{\gamma_2(A)}\Big]
\subseteq \!\sqrt{\gamma_{n+2}(A)} , \\
[A, [\gamma_{n}(A),C]] &\subseteq \Big[A, \Big[\!\sqrt{\gamma_{n}(A)},C\Big]\Big]\subseteq 
\Big[A, \!\sqrt{\gamma_{n+1}(A)}\Big]\subseteq \!\sqrt{\gamma_{n+2}(A)} .
\end{align*} 
Thus, by Lemma \ref{lem:3subgroups}, 
$[\gamma_{n+1}(A),C]=[[A,\gamma_n(A)],C]$ is a subgroup of 
$\!\sqrt{\gamma_{n+2}(A)}$. 
Hence, $[c,a]^{k} \in \!\sqrt{\gamma_{n+2}(A)}$, and so  
$[c,a]\in \!\sqrt{\gamma_{n+2}(A)}$. This shows that 
$\big[\!\sqrt{\gamma_{n+1}(A)},C\big]\subseteq \!\sqrt{\gamma_{n+2}(A)}$, 
and the induction on $n$ is complete. 

Assume now that $\big[\!\sqrt{\gamma_n(A)}, \gamma_{m}(C)\big]\subseteq 
\!\sqrt{\gamma_{m+n}(A)}$ for all $n\ge 1$. 
Let $a \in \!\sqrt{\gamma_{n+1}(A)}$, so that $a^k \in \gamma_{n+1}(A)$, for 
some $k>0$, and let $c\in \gamma_m(C)$. 
Since $a$ belongs to $\!\sqrt{\gamma_{n}(A)}$, too, 
we may apply Lemma \ref{lem:conj-mod} to deduce that 
\[
[c,a]^{k} \equiv [c,a^{k}] \:\:\operatorname{ modulo } 
\Big[\!\sqrt{\gamma_{n}(A)}, \Big[\!\sqrt{\gamma_{n}(A)},\gamma_m(C)\Big] \Big] .
\] 
By the induction hypothesis, this subgroup is contained in 
$[\!\sqrt{\gamma_{n}(A)}, \!\sqrt{\gamma_{m+n}(A)}]$, 
which in turn is a subgroup of 
$\!\sqrt{\gamma_{m+2n}(A)} \subseteq \!\sqrt{\gamma_{m+n+1}(A)}$,  
since $\!\sqrt{\gamma(A)}$ is an $N$-series and $n\ge 1$.  On the 
other hand, $[c,a^k]\in [\gamma_{n+1}(A),\gamma_m(C)]\subseteq 
 [\!\sqrt{\gamma_{n+1}(A)},\gamma_m(C)]$, which is a 
subgroup of $\!\sqrt{\gamma_{m+n+1}(A)}$, again by the induction 
hypothesis. Hence, $[c,a]^{k} \in \!\sqrt{\gamma_{m+n+1}(A)}$, and so  
$[c,a]\in \!\sqrt{\gamma_{m+n+1}(A)}$. 
We have thus shown that 
$[\!\sqrt{\gamma_n(A)}, \gamma_{m+1}(C)]$ is contained 
in $\!\sqrt{\gamma_{m+n+1}(A)}$, 
and the induction on $m$ is complete. 
This ends the proof.
\end{proof}
 
We are now ready to state and prove the main result of this section.

\begin{theorem}
\label{theorem:fr-rational}
Let $B=A\rtimes_{\varphi} C$ be a split extension of groups, and 
let $L=\{L_n\}_{n\ge 1}$ be the corresponding $N$-series for $A$. 
Suppose $C$ acts trivially on $A_{\abf}$. Then 
\begin{enumerate}[itemsep=3pt]
\item $\!\sqrt[A]{L_n}=\!\sqrt[A]{\gamma_n(A)}$ for each $n\ge 1$.
\item $\!\sqrt[B]{\gamma_n(B)}=
\!\sqrt[A]{\gamma_n(A)}\rtimes_{\varphi} \!\sqrt[C]{\gamma_n(C)}$.
\end{enumerate}
\end{theorem}

\begin{proof}
By Lemma \ref{lem:akb}, $\gamma_n(A)\subseteq L_n$, and thus 
$\!\sqrt{\gamma_n(A)} \subseteq \!\sqrt{L_n}$.  We prove the 
reverse inclusion by induction on $n$. Since $L_1=A$, 
the base case $n=1$ is immediate.  So assume that 
$\!\sqrt{L_{n}}\subseteq\!\sqrt{\gamma_{n}(A)}$. 
Recall from \eqref{eq:sqrt-ln} that 
every element in $\!\sqrt{L_{n+1}}$ admits a non-trivial power which 
can be expressed as a word in elements  from the subgroups 
$[A,L_{n}]$, $[A,\gamma_{n}(C)]$, and $[L_{n},C]$. It is enough, 
then, to show that each of these subgroups is contained in 
$\!\sqrt{\gamma_{n+1}(A)}$. We consider the three cases in turn.

\begin{enumerate}[itemsep=4pt, topsep=3pt, partopsep=3pt, label=(\alph*)]
\item \label{aa}
By the induction hypothesis, 
we have that $[A,L_{n}]\subseteq [A,\!\sqrt{L_{n}}] \subseteq 
[A, \!\sqrt{\gamma_{n}(A)}]\subseteq 
\!\sqrt{\gamma_{n+1}(A)}$.

\item \label{bb}
By the induction hypothesis and Lemma \ref{lem:act-sqrt}, 
we have that $[L_{n},C]\subseteq [\!\sqrt{L_{n}},C] \subseteq 
[\!\sqrt{\gamma_{n}(A)},C]\subseteq 
\!\sqrt{\gamma_{n+1}(A)}$.

\item \label{cc}
By Lemma \ref{lem:act-sqrt} again, 
we have that $[A,\gamma_{n}(C)]\subseteq 
\!\sqrt{\gamma_{n+1}(A)}$.
\end{enumerate}

This shows that $\!\sqrt{L_{n+1}} \subseteq \!\sqrt{\gamma_{n+1}(A)}$, 
thus completing the induction, and finishing the proof of the first claim.
The second claim now follows from the first one and Theorem \ref{thm:gp-rat}.
\end{proof}

\begin{corollary}
\label{cor:res-tf-nilp}
Let $B=A\rtimes C$ be a split extension of RTFN groups. If $C$ acts 
trivially on $A_{\abf}$, then $B$ is also RTFN.
\end{corollary}

\begin{proof}
Recall from Lemma \ref{lem:rtfn} that a group $G$ is 
residually torsion-free nilpotent (RTFN) precisely when 
$\gamma^{\rat}_{\omega}(G)=\{1\}$. 
The claim follows from Theorem \ref{theorem:fr-rational} 
by an argument similar to the one employed in the 
proof of Corollary \ref{cor:res-nilp}.
\end{proof}

As a consequence of Theorem \ref{theorem:fr-rational}, we also obtain 
the following rational analogue of Falk and Randell's theorem.

\begin{theorem}
\label{thm:q-fr}
Let $\begin{tikzcd}[column sep=16pt]
1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\end{tikzcd}$ 
be a split exact sequence of groups, with monodromy 
$\varphi\colon C\to \Aut(A)$. 
Suppose $C$ acts trivially on $A_{\abf}$. Then 
\begin{equation}
\label{eq:ses-rat}
\begin{tikzcd}[column sep=32pt]
0\ar[r] &  \gr^{\rat}(A) \ar[r, "\gr^{\rat}(\alpha)"] &
\gr^{\rat}(B) \ar[r, "\gr^{\rat}(\beta)"] & \gr^{\rat}(C) \ar[r] & 0
\end{tikzcd}
\end{equation}
is a split exact sequence of graded Lie algebras, 
and so $\gr^{\rat}(B) \cong 
\gr^{\rat}(A) \rtimes_{\bar\varphi} \gr^{\rat}(C)$, where 
$\bar\varphi\colon \gr^{\rat}(C) \to \Der (\gr^{\rat}(A))$ 
is the map of Lie algebras induced by $\varphi$.
\end{theorem} 

\begin{example}
\label{ex:klein-rat-bis}
Let $K= \langle a,t \mid tat^{-1} =a^{-1}\rangle$ be the 
Klein bottle group, and consider the semidirect product 
$G=K\rtimes_{\varphi} \mathbb{Z}$ with monodromy action given by 
$t\mapsto ta$ and $a\mapsto a$. Clearly, $\mathbb{Z}$ acts trivially 
on $K_{\abf}$ (though not on $K_{\ab}$).  Thus, we may 
apply Theorem \ref{thm:q-fr}; using the computation 
from Example \ref{ex:klein-rat}, we conclude that 
$\gr^{\rat}(G)=\mathbb{Z}^2$, concentrated in degree $1$. 
\end{example}

\section{The mod-$p$ lower central series of a split extension}
\label{sect:p-split}

\subsection{A $p$-torsion $N$-series}
\label{subsec:split-n-series-p}

Given a split extension of groups, $B=A\rtimes_{\varphi} C$, 
and a prime $p$, we define by analogy with \eqref{eq:subgroup}  a 
sequence of subgroups of $A$, denoted $\big\{L^{p}_n\big\}_{n\ge 1}$, 
as follows. We set $L^{p}_1=A$ and define the other terms inductively by 
\begin{equation}
\label{eq:subgroup-p}
L^{p}_{n+1}=\big\langle \big(L^{p}_{n}\big)^p, \big[A,L^{p}_{n}\big], 
\big[A, \gamma^{p}_n(C)\big], \big[L^{p}_{n},C\big]\big\rangle \, .
\end{equation}

\begin{lemma}
\label{lem:gp-normal-p} 
The groups $L^{p}_n$ are normal subgroups of $A$.
\end{lemma}

\begin{proof}
We establish the  claim by induction on $n$. The base case 
$L^{p}_1=A$ is tautologically true. So assume $L^{p}_n \triangleleft A$.  
Since $ax^pa^{-1}=(axa^{-1})^p$, this immediately implies that 
$\big(L^{p}_n\big)^p \triangleleft A$. Since commutators of normal 
subgroup are again normal, $[A,L^{p}_n]  \triangleleft A$; moreover, 
by Lemma \ref{lem:normal}, $[A,\gamma^{p}_{n}(C)]  \triangleleft A$. 
Finally, if $x\in A$, then $x[L^{p}_n,C] x^{-1} \subseteq  
[L^{p}_n,C] \cdot [A,L^{p}_n]$, by formula \eqref{eq:conj-2}.  
Therefore, $L^{p}_{n+1}\triangleleft A$, and the induction is complete.
\end{proof}

\begin{lemma}
\label{lem:gp-series-p} 
The subgroups $\{L^{p}_n\}_{n\ge 1}$ form a descending series for $A$. 
\end{lemma}

\begin{proof}
We need to show that 
$L^{p}_{n+1} \subseteq L^{p}_{n}$ for all $n\ge 1$ or, equivalently, 
that $\big(L^{p}_{n}\big)^p$, $[A, L^{p}_n]$, $[A,\gamma^{p}_n(C)]$, 
and $[L^{p}_n,C]$ are all included in $L^{p}_n$.  

\begin{enumerate}[itemsep=4pt, topsep=3pt, partopsep=3pt, label=(\alph*)]
\item \label{dd0-p}
$\big(L^{p}_{n}\big)^p \subseteq L^{p}_n$: this is obvious.

\item \label{dd1-p}
$[A, L^{p}_n]\subseteq L^{p}_n$: follows at once from 
Lemma \ref{lem:gp-normal-p}.

\item  \label{dd2-p}
$[A,\gamma^{p}_n(C)]\subseteq L^{p}_n$: we prove this by induction 
on $n$, with the base case, $[A,C]\subseteq A$, being obvious. We have 
$[[A,C],\gamma^{p}_n(C)]\subseteq [A,\gamma^{p}_n(C)] \subseteq L^{p}_{n+1}$, 
by \eqref{eq:subgroup-p} and 
$[[A,\gamma^{p}_n(C)], C]\subseteq [L^{p}_n,C] \subseteq L^{p}_{n+1}$, 
by the induction hypothesis and \eqref{eq:subgroup-p}. 
Thus,  $[A,[C,\gamma^{p}_n(C)]]\subseteq L^{p}_{n+1}$, 
by Lemma \ref{lem:3subgroups}.
By \eqref{eq:subgroup-p}, we also have 
$\big[A, \big(\gamma^{p}_n(C)\big)^{p}\big] \subseteq 
[A, \gamma^{p}_n(C) ]  \subseteq  L^{p}_{n+1}$.
Hence, $[A,\gamma^{p}_{n+1}(C)] \subseteq L^{p}_{n+1}$, 
and the induction is complete.

\item  \label{dd3-p}
$[L^{p}_n,C] \subseteq L^{p}_n$: we prove this by induction on $n$, with 
the base case $n=1$ being obvious. For the induction step, 
there are four cases to consider. 
\begin{enumerate}[itemsep=3pt, topsep=3pt, partopsep=3pt, label=(\roman*)]
\item \label{la0-p}
Clearly, $[ \big(L^{p}_{n}\big)^p , C]\subseteq 
[ L^{p}_{n} , C]\subseteq  L^{p}_{n+1}$.
\item \label{la1-p}
Clearly, $\big[[A,C], L^{p}_n\big]\subseteq [A,L^{p}_n]$, while  
$[A,[L^{p}_n, C]]\subseteq  [A,L^{p}_n]$ by the induction hypothesis. 
Therefore, $[[A,L^{p}_n], C]\subseteq [A,L^{p}_n] \subseteq  L^{p}_{n+1}$, 
by Lemma \ref{lem:3subgroups} and \eqref{eq:subgroup-p}.
\item \label{la2-p}
We have $[[A,C],\gamma^{p}_n(C)]\subseteq [A,\gamma^{p}_n(C)]\subseteq L^{p}_{n+1}$, 
by \eqref{eq:subgroup-p}, 
and $[A,[C,\gamma^{p}_n(C)]] \subseteq 
[A,\gamma^{p}_{n+1}(C)]\subseteq L^{p}_{n+1}$,  
by case \ref{dd2-p}.  Therefore, 
$[[A,\gamma^{p}_n(C)], C]\subseteq L^{p}_{n+1}$, 
by Lemma \ref{lem:3subgroups}.
\item \label{la3-p}
Finally, $[[L^{p}_n,C],C]\subseteq [L^{p}_n,C] \subseteq L^{p}_{n+1}$, 
by the induction hypothesis and \eqref{eq:subgroup-p}.
\end{enumerate}
This shows  that $[L^{p}_{n+1} ,C]\subseteq L^{p}_{n+1}$, thereby 
completing the induction.
\end{enumerate}
This ends the proof.
\end{proof}

\begin{lemma}
\label{lem:lpc} 
$\big[L^{p}_{n},\gamma^{p}_{m}(C)\big]\subseteq L^{p}_{n+m}$, 
for all $n,m \ge 1$.
\end{lemma}

\begin{proof}
We establish the  claim by induction on $m$. The base case $m=1$, 
which amounts to $[L^{p}_n, C]\subseteq L^{p}_{n+1}$ for all $n\ge 1$, 
follows directly from \eqref{eq:subgroup-p}. Assume now that 
$[L^{p}_{n},\gamma_{m}(C)]\subseteq L^{p}_{n+m}$, for all $n\ge 1$.
We then have $[[L^{p}_{n},C],\gamma^{p}_{m}(C)] 
\subseteq [L^{p}_{n+1},\gamma^{p}_{m}(C] \subseteq L^{p}_{n+m+1}$ 
and $ [[L^{p}_{n},\gamma^{p}_{m}(C)],C] 
\subseteq [L^{p}_{n+m},C] \subseteq  L^{p}_{n+m+1}$, and so  
$[L^{p}_{n},[C,\gamma^{p}_{m}(C)] ]$ is 
contained in $L^{p}_{n+m+1}$, by Lemma \ref{lem:3subgroups}.
We also have $[L^{p}_{n},\big(\gamma^{p}_{m}(C)\big)^p] 
\subseteq [L^{p}_{n}, \gamma^{p}_{m+1}(C)] \subseteq L^{p}_{n+m+1}$. 
Therefore, $[L^{p}_{n},\gamma^{p}_{m+1}(C)]$ is 
contained in $L^{p}_{n+m+1}$, and so  
the induction step is complete.
\end{proof}

\begin{theorem}
\label{thm:p-n-seq}
The sequence of subgroups $\big\{L^{p}_n\big\}_{n\ge 1}$ forms a $p$-torsion 
$N$-series for $A$.
\end{theorem}

\begin{proof}
From the way the sequence was defined in \eqref{eq:subgroup-p}, we 
have that $\big(L^{p}_{n}\big)^p\subseteq L^{p}_{n+1}$ for all $n\ge 1$;  
thus, $L^{p}$ is a $p$-torsion series.

Next, we show that $L^{p}$ is an $N$-series, that is, 
$[L^{p}_n, L^{p}_m]\subseteq L^{p}_{n+m}$ for all $n,m\ge 1$. 
We prove this claim by induction on $m$. The base case $m=1$, 
which amounts to $[L^{p}_n, A]\subseteq L^{p}_{n+1}$, 
follows at once from definition \eqref{eq:subgroup-p}.  Assume now that 
$[L^{p}_n, L^{p}_m]\subseteq L^{p}_{n+m}$ for all $n\ge 1$.
To prove the induction step, there are four cases to consider.

\begin{enumerate}[itemsep=4pt, topsep=3pt, partopsep=3pt, label=(\alph*)]
\item \label{pz0}
$[L^{p}_n, \big(L^{p}_{m}\big)^p] \subseteq L^{p}_{n+m+1}$: 
to prove this assertion, 
consider elements $a\in L^{p}_n$ and $b\in L^{p}_{m}$. 
Applying Lemma \ref{lem:conj-mod} shows that 
$[a,b^p] \equiv [a,b]^p$ modulo $[[L^{p}_n,L^{p}_{m}],L^{p}_{m}]$, 
a subgroup of $L^{p}_{n+2m}\subseteq L^{p}_{n+m+1}$. 
But $[a,b]\in [L^{p}_n, L^{p}_m] \subseteq  
L^{p}_{n+m}$; thus,  $[a,b]^p \in 
\big( L^{p}_{n+m} \big)^p \subseteq L^{p}_{n+m+1}$, 
showing that $[a,b^p]\in L^{p}_{n+m+1}$.
\item \label{pz1}
$[[L^{p}_{n}, A],L^{p}_{m}] \subseteq [L^{p}_{n+1},L^{p}_m] \subseteq L^{p}_{n+1+m}$ 
and $ [[L^{p}_{n}, L^{p}_m],A] \subseteq [L^{p}_{n+m},A]\subseteq L^{p}_{n+1+m}$. 
Therefore, $[L^{p}_n,[A,L^{p}_{m}]]\subseteq L^{p}_{n+1+m}$. 
\item \label{pz2}
$[[L^{p}_n, A], \gamma^{p}_{m}(C)] \subseteq [L^{p}_{n+1},\gamma^{p}_{m}(C)] 
\subseteq L^{p}_{n+1+m}$ and $[[L^{p}_n, \gamma^{p}_{m}(C)], A]
\subseteq [L^{p}_{n+m},A] \subseteq L^{p}_{n+1+m}$. 
Therefore, $[L^{p}_n,[A, \gamma^{p}_{m}(C)]]  \subseteq L^{p}_{n+1+m}$.
\item \label{pz3}
$[[L^{p}_{n}, C],L^{p}_{m}] \subseteq [L^{p}_{n+1},L^{p}_{m}] \subseteq L^{p}_{n+1+m}$
and $ [[L^{p}_{n}, L^{p}_m],C] \subseteq [L^{p}_{n+m},C] \subseteq L^{p}_{n+1+m}$. 
Therefore, $[L^{p}_{n}, [L^{p}_{m},C]] \subseteq L^{p}_{n+1+m}$.
\end{enumerate}
This shows that $[L^{p}_n,L^{p}_{m+1}]\subseteq L^{p}_{n+m+1}$ for all $n\ge 1$, 
thereby completing the induction.
\end{proof}

We conclude this subsection with one more lemma. 

\begin{lemma}
\label{lem:akb-p} 
The inclusions $\gamma^{p}_n(A) \subseteq L^{p}_n\subseteq \gamma^{p}_n(B)$ 
hold for all $n\ge 1$.
\end{lemma}

\begin{proof}
The first claim follows at once from Lemma \ref{lem:fastest-p} and 
Theorem \ref{thm:p-n-seq}. The second claim is proved 
by induction on $n$, with the base case $n=1$ being obvious. 
Assuming $L^{p}_{n}\subseteq \gamma^{p}_{n}(B)$, we have 
that all four subgroups generating $L^{p}_{n+1}$ in \eqref{eq:subgroup-p} 
are included in either $\big(\gamma^{p}_{n}(B)\big)^p$ or 
$[B,\gamma^{p}_{n}(B)]$, whence 
$L^{p}_{n+1}\subseteq \gamma^{p}_{n+1}(B)$.
\end{proof}

\subsection{Split extensions and the mod-$p$ LCS}
\label{subsec:p-fr}

We are now in a position to state and prove the main result of this section.

\begin{theorem}
\label{thm:gp-p}
Let $B=A\rtimes_{\varphi} C$ be a split extension. 
For each $n\ge 1$, the homomorphism $\varphi\colon C\to \Aut(A)$ 
restricts to a homomorphism  $\varphi\colon \gamma^{p}_n(C)\to 
\Aut\big(L^{p}_n\big)$.  Moreover,  
\begin{equation}
\label{eq:spli-lp}
\gamma^{p}_n(B)=L^{p}_n\rtimes_{\varphi} \gamma^{p}_n(C)\, .
\end{equation}
\end{theorem}

\begin{proof}
We prove the first claim by induction on $n$, the case $n=1$ being tautological. 
So assume the map $\varphi\colon C\to \Aut(A)$ restricts to a map 
$\varphi\colon \gamma^{p}_{n-1}(C)\to \Aut(L^{p}_{n-1})$.
Let  $g$ be in $\gamma^{p}_{n}(C)$. For the induction step, 
there are four cases to consider. 

\begin{enumerate}[(a)]
\item \label{000-p}
Clearly, $\varphi(g)$ leaves $\big(L^{p}_{n-1}\big)^p$ invariant.

\item \label{aaa-p}
Since $g\in \gamma^{p}_{n-1}(C)$, too, $\varphi(g)$ leaves 
$L^{p}_{n-1}$ invariant. Formula \eqref{eq:act-1} now implies that 
$\varphi(g)$ leaves $[A,L^{p}_{n-1}]$ invariant.
\item \label{bbb-p}
Since $gc\in \gamma^{p}_{n-1}(C)$ for $c\in \gamma^{p}_{n-1}(C)$, 
formula \eqref{eq:act-2} implies that $\varphi(g)$ leaves 
$[A,\gamma^{p}_{n-1}(C)]$ invariant.
\item \label{ccc-p}
By Lemma \ref{lem:normal}, $[L^{p}_{n-1},C]$ is a normal subgroup of $A$. 
Formula \eqref{eq:act-2}  now implies that $\varphi(g)$ leaves 
$[L^{p}_{n-1},C]$ invariant.
\end{enumerate}
By definition, $L^{p}_n$ is generated by the subgroups considered in the four 
cases above; hence, $\varphi(g)$ leaves  $L^{p}_n$ invariant, too. 

Since the map $\varphi(g)\colon A\to A$ is injective, its restriction to $L^{p}_n$ is 
also injective. To show that the map $\varphi(g)\colon L^{p}_n\to L^{p}_n$ 
is surjective, we go through the above four cases one more time. 
In case \ref{000-p}, the fact that the map  
$\varphi(g)\colon L^{p}_{n-1}\to L^{p}_{n-1}$ is surjective 
implies that the map  
$\varphi(g)\colon \big(L^{p}_{n-1}\big)^p\to \big(L^{p}_{n-1}\big)^p$ 
is also surjective. 
In case \ref{aaa-p}, the surjectivity of the maps  $\varphi(g)\colon A\to A$  
and $\varphi(g)\colon L^{p}_{n-1}\to L^{p}_{n-1}$ yield the claim. 
In cases \ref{bbb-p} and \ref{ccc-p} the claim follows from \eqref{eq:act-2}, by 
observing that $[c,a]=\varphi(g)([g^{-1}c,a] \cdot  {}^a[g^{-1},a^{-1}])$.

The second claim is also proved by induction on $n$, 
with the case $n=1$ being tautological. Assume that 
$\gamma_{n-1}(B)=L^{p}_{n-1}\rtimes_{\varphi} \gamma^{p}_{n-1}(C)$. 
To show that $L^{p}_{n}\rtimes_{\varphi} \gamma^{p}_{n}(C)
\subseteq \gamma^{p}_{n}(B)$, observe that 
$\gamma^{p}_n(C)\subseteq \gamma^{p}_n(B)$,  
and also $L^{p}_n\subseteq \gamma^{p}_n(B)$, by Lemma \ref{lem:akb-p}. 
For the reverse inclusion, the induction hypothesis and 
definition \eqref{eq:subgroup-p} show that
\begin{equation}
\label{eq:gamma-n-b-p}
\gamma^{p}_n(B)=[B,\gamma^{p}_{n-1}(B)]\subseteq 
[A\rtimes_{\varphi}  C,L^{p}_{n-1}\rtimes_{\varphi} \gamma^{p}_{n-1}(C)]
\subseteq L^{p}_n \rtimes_{\varphi} \gamma^{p}_n(C) \,.
\end{equation}
This completes the proof.
\end{proof}

Let $\gr^{L^{p}}(A)$ be the graded Lie algebra associated to the $N$-series 
$L^{p}=\big\{L^{p}_n\big\}_{n\ge 1}$. Since $L^{p}$ is a 
$p$-torsion series, the graded pieces of this Lie algebra are 
$\mathbb{Z}_p$-vector spaces. 

\begin{corollary}
\label{cor:gp-gr-p}
Let $\begin{tikzcd}[column sep=16pt]
\! 1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\!\end{tikzcd}$ 
be a split exact sequence of groups.  For each prime $p$, 
there is an induced split exact sequence of graded Lie algebras, 
\begin{equation}
\label{eq:split-gr-p}
\begin{tikzcd}[column sep=20pt]
\!\!0\ar[r] &  \gr^{L^{p}}(A) \ar[rr, "\gr^{L^{p}\!}(\alpha)"] &&
\gr^p(B) \ar[rr, "\gr(\beta)"] && \gr^p(C) \ar[r] & 0\, .
\end{tikzcd}
\end{equation}
Consequently,  $\gr^{p}(B) \cong 
\gr^{{L^p}}(A) \rtimes_{\bar\varphi} \gr^{p}(C)$, where 
$\bar\varphi\colon \gr^{p}(C) \to \Der (\gr^{{L^p}}(A))$ 
is the map of Lie algebras induced by the monodromy 
$\varphi\colon C\to \Aut(A)$ of the extension.
\end{corollary} 

\begin{proof}
The proof is similar to that of Theorem \ref{thm:gp-gr}.
\end{proof}

\section{Trivial action on $\mathbb{Z}_p$-homology}
\label{sect:modp-fr}

\subsection{Mod-$p$ almost direct products}
\label{subsec:p-bg}
Following \cite{BG}, we say that a split extension $B=A\rtimes_{\varphi} C$ 
is a {\em mod-$p$ almost direct product}\/ if the group $C$ 
acts trivially on $A_{\ab}\otimes \mathbb{Z}_p=H_1(A,\mathbb{Z}_p)$. 
This condition may be expressed as 
$\varphi(c)(a) \cdot a^{-1}  \in \gamma_2^p(A)$, for all $c\in C$ 
and $a\in A$. Alternatively, if we view $C$ as a subgroup of 
$G$ via the splitting $\sigma\colon C\to B$, the condition simply reads
\begin{equation}
\label{eq:ac-gamma-p}
[A,C]\subseteq \gamma^p_2(A)\,.
\end{equation}

As noted in \cite[Proposition 3.2]{BG}, the fact that a split extension 
$B=A\rtimes_{\varphi} C$ is a mod-$p$ almost direct product 
does not depend on the choice of splitting $\sigma$.

\begin{lemma}
\label{lem:act-modp}
Assume that $[A,C]\subseteq \gamma^p_2(A)$.
Then, for all $m, n\ge 1$,
\begin{equation}
\label{eq:gamma-ac-modp}
\big[ \gamma^p_n(A), \gamma^p_m(C)\big]\subseteq 
\gamma^p_{m+n}(A).
\end{equation}
\end{lemma}

\begin{proof}
We prove the claim by induction on $m$. The case $m=1$, 
which amounts to $[\gamma^p_n(A),C]\subseteq \gamma^p_{n+1}(A)$ 
for all $n\ge 1$, is proved by induction on $n$. The base case, 
$[A,C]\subseteq \gamma^p_2(A)$, is our assumption. 
So assume that $[\gamma^p_{n}(A),C]\subseteq \gamma^p_{n+1}(A)$. 
For the induction step, we need to show that 
$[[A,\gamma^p_n(A)],C]$ and $[(\gamma^p_n(A))^p,C]$ 
are both included in $\gamma^p_{n+2}(A)$.

\begin{enumerate}[(a)]
\item \label{tp1}
Our assumption, the induction hypothesis, and the fact that $\gamma(A)$ 
is an $N$-series imply that 
$[\gamma^p_{n}(A), [A,C]] \subseteq 
[\gamma^p_{n}(A), \gamma^p_2(A)]
\subseteq \gamma^p_{n+2}(A)$ and 
$[A, [\gamma^p_{n}(A),C]] \subseteq 
[A, \gamma^p_{n+1}(A)]\subseteq \gamma^p_{n+2}(A)$. 
The first inclusion now follows from Lemma \ref{lem:3subgroups}.

\item \label{tp2}
Next, let $a\in \gamma^{p}_n(A)$ and $c\in C$. 
Applying Lemma \ref{lem:conj-mod} shows that 
$[c,a^p] \equiv [c,a]^p$ modulo 
$[\gamma^{p}_n(A), [\gamma^{p}_n(A),C]]$, 
a subgroup of $[\gamma^{p}_n(A), \gamma^{p}_{n+1}(A)] 
\subseteq \gamma^{p}_{2n+1}(A) \subseteq \gamma^{p}_{n+2}(A) $. 
But $[c,a]\in [\gamma^p_n(A),C] \subseteq  
\gamma^{p}_{n+1}(A)$, and so $[c,a]^p\in   
\big(\gamma^{p}_{n+1}(A)\big)^p\subseteq  
\gamma^{p}_{n+2}(A)$. Thus,  $[c,a^p] \in \gamma^{p}_{n+2}(A)$, 
and the second inclusion is also proved.
\end{enumerate}

Assume now that $[\gamma^p_n(A), \gamma^p_{m}(C)]\subseteq 
\gamma^p_{m+n}(A)$ for all $n\ge 1$. For the induction step, we need 
to show that $[\gamma^p_n(A),[\gamma^p_{m}(C),C]]$ and 
$[(\gamma^p_n(A))^p,(\gamma^p_{m}(C))^p]$ 
are both included in $\gamma^p_{m+n+1}(A)$.

\begin{enumerate}[(a)]
\item \label{tpp1}
The previous induction (on $n$) and our current induction 
hypothesis (on $m$) give 
$[[\gamma^p_{n}(A), C],\gamma^p_{m}(C]] \subseteq 
[\gamma^p_{n+1}(A), \gamma^p_{m}(C)]
\subseteq \gamma^p_{m+n+1}(A)$ and 
$[[\gamma^p_{n}(A),\gamma^p_{m}(C)],C]  \subseteq 
[\gamma^p_{m+n}(A),C]\subseteq \gamma^p_{m+n+1}(A)$. 
The first inclusion now follows from Lemma \ref{lem:3subgroups}.

\item \label{tpp2}
Next, let $a\in \gamma^{p}_n(A)$ and $c\in \gamma^p_m(C)$. 
By Lemma \ref{lem:conj-mod}, we have  that 
$[a,c^p] \equiv [a,c]^p$ modulo 
$[[\gamma^{p}_n(A), \gamma^{p}_m(C)],\gamma^p_{m}(C)]]$, 
a subgroup of $[\gamma^{p}_{m+n}(A), \gamma^{p}_{m}(C)] 
\subseteq \gamma^{p}_{2m+n}(A) \subseteq \gamma^{p}_{m+n+1}(A) $. 
But $[a,c]\in [\gamma^p_n(A),\gamma^p_m(C)] \subseteq  
\gamma^{p}_{m+n}(A)$, and so $[a,c]^p\in   
\big(\gamma^{p}_{m+n}(A)\big)^p\subseteq  
\gamma^{p}_{m+n+1}(A)$. Thus,  $[a,c^p] \in 
\gamma^{p}_{m+n+1}(A)$, and the second inclusion 
is proved as well.
\end{enumerate}
This completes the proof.
\end{proof}

\begin{theorem}
\label{thm:fr-lem-p}
Let $B=A\rtimes_{\varphi} C$ be a mod-$p$ almost direct product. 
Then $L^{p}_n=\gamma^{p}_n(A)$ for all $n\ge 1$.
\end{theorem}

\begin{proof}
By Lemma \ref{lem:akb-p}, we have that $\gamma^{p}_n(A)\subseteq L^{p}_n$, 
for all $n\ge 1$. We prove the reverse inclusion by induction on $n$, with the base 
case $n=1$ being obvious.  Assume $L^{p}_n \subseteq \gamma^{p}_n(A)$, 
so that, in fact, $L^{p}_n = \gamma^{p}_n(A)$.
For the induction step, it is enough to show that all four subgroups 
from \eqref{eq:subgroup-p} that generate $L^p_{n+1}$ are included 
in $\gamma^{p}_{n+1}(A)$. 

By the definition of $\gamma^{p}(A)$, we have that 
$\big(L^{p}_{n}\big)^p = \big(\gamma^{p}_{n}(A)\big)^p  
\subseteq \gamma^{p}_{n+1}(A)$ and 
$[A,L^{p}_{n}]=[A,\gamma^{p}_{n}(A)] \subseteq \gamma^{p}_{n+1}(A)$. 
Moreover, by Lemma \ref{lem:act-modp}, 
$[A,\gamma^{p}_{n}(C)]\subseteq \gamma^{p}_{n+1}(A)$ and 
$[L^{p}_{n},C]=[\gamma^{p}_{n}(A),C]\subseteq \gamma^{p}_{n+1}(A)$. 
This finishes the induction and completes the proof.
\end{proof}

\subsection{Applications}
\label{subsec:almost-direct-fr-p}
In \cite{BG}, Bellingeri and Gervais proved mod-$p$ versions of two 
of the Falk and Randell theorems.  We recover their result by our method. 

\begin{theorem}[\cite{BG}]
\label{thm:fr-p}
Let $B=A\rtimes_{\varphi} C$ be a mod-$p$ almost direct product. Then,
\begin{enumerate}[topsep=3pt, itemsep=3pt]
\item \label{fr1-p}
$\gamma^{p}_n(B)=\gamma^{p}_n(A)\rtimes_{\varphi} \gamma^{p}_n(C)$, 
for all $n\ge 1$.

\item \label{fr2-p}
If $A$ and $C$ are residually $p$-finite, then $B$ is also 
residually $p$-finite.
\end{enumerate}
\end{theorem} 

\begin{proof}
The first claim follows at once from Theorems \ref{thm:gp-p} and \ref{thm:fr-lem-p}.
The second claim follows from claim \eqref{fr1-p} and Lemma \ref{lem:paris-bis} by an 
argument entirely similar to the one given in the proof of Corollary \ref{cor:res-nilp}.
\end{proof}

We conclude with a mod-$p$ analogue of Falk and Randell's theorem 
on the decomposition of the associated graded Lie algebra of an almost 
direct product.

\begin{theorem}
\label{thm:p-fr}
Let $\begin{tikzcd}[column sep=16pt]
1\ar[r] &  A \ar[r, "\alpha"]
& B \ar[r, "\beta"] & C \ar[r] & 1\end{tikzcd}$ 
be a split exact sequence of groups, with monodromy 
$\varphi\colon C\to \Aut(A)$.  Suppose $C$ acts trivially 
on $H_1(A,\mathbb{Z}_p)$. Then the sequence
\begin{equation}
\label{eq:ses-p}
\begin{tikzcd}[column sep=20pt]
0\ar[r] &  \gr^{p}(A) \ar[rr, "\gr^{p}(\alpha)"] &&
\gr^{p}(B) \ar[rr, "\gr^{p}(\beta)"] && \gr^{p}(C) \ar[r] & 0\, ,
\end{tikzcd}
\end{equation}
is a split exact sequence of graded Lie algebras; thus, $\gr^{p}(B) \cong 
\gr^{p}(A) \rtimes_{\bar\varphi} \gr^{p}(C)$, where 
$\bar\varphi\colon \gr^{p}(C) \to \Der (\gr^{p}(A))$ 
is the map of Lie algebras induced by $\varphi$.
\end{theorem} 

\begin{proof}
By Theorem \ref{thm:fr-lem-p}, we have that $L^p_n=\gamma^p_n(A)$ for all $n\ge 1$, 
and so $\gr^{L^p}(A)=\gr^p(A)$. The claim now follows from Corollary \ref{cor:gp-gr-p}.
\end{proof}

\begin{example}
\label{ex:klein-2}
Let $G=A\rtimes C= \langle a,t \mid tat^{-1} =a^{-1}\rangle$ be the 
Klein bottle group from Examples \ref{ex:klein} and \ref{ex:klein-rat}. 
Note that $C=\langle t\rangle$ acts trivially on $H_1(A,\mathbb{Z}_2)=\mathbb{Z}_2$, 
and thus Theorem \ref{thm:p-fr} applies, with $p=2$.
From Example \ref{ex:z-p}, we know that 
$\gr^2_n(A)=\mathbb{Z}_2=\langle a^{2^{n-1}}\rangle$ and 
$\gr^2_n(C)=\mathbb{Z}_2=\langle t^{2^{n-1}}\rangle$;  
therefore, $\gr^2_n(G)=\mathbb{Z}_2\oplus \mathbb{Z}_2$ for all $n\ge 1$.
Yet the extension \eqref{eq:ses-p} is nontrivial in this case, 
since $[t,a]=a^2$.  Interestingly, $\gr^2(G)\not\cong \gr^2(\mathbb{Z}^2)$, 
although $H^*(G,\mathbb{Z}_2)\cong H^*(\mathbb{Z}^2,\mathbb{Z}_2)$, as graded rings. 
\end{example}


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